No, it isn't. Firstly you've changed the equation, the equation which I originally corrected you on was, according to you, $$\omega = \alpha k + M$$, where you've mixed scalars and matrices. Secondly $$E=p+M$$ isn't true in relativity. $$-E^{2} + p^{2} = -m^{2}$$ can be rearranged to $$E^{2} = m^{2} + p^{2}$$ but that doesn't imply $$E=m+p$$. That is
precisely the mistake you made in the thread in question, one I spent considerable time explaining to you. You complain I waffled too much but you've just shown you didn't even learn anything from it!
In fact, I explained at length
here why you can't say $$E = \sqrt{p^{2}+M^{2}} = p+M$$ but it's something you could
mistakenly conclude given the structure of the Dirac equation. It's a structure the whole $$\{ \gamma^{\mu} , \gamma^{\nu} \} = 2g^{\mu\nu}\mathbb{I}$$ thing we talked about is for!
I'll explain it
again since obviously you didn't understand it before, you didn't understand the explanation and you still don't understand it.
Oscillating things obey the massive wave equation, which for v=1 is $$(-\partial_{t}^{2} + \nabla^{2})A = M^{2}A$$. This is second order as there's second order derivatives. Got that or do I need to go over it again? If you Fourier transform this or hit it on $$e^{i(\omega t - \mathbf{k} \cdot \mathbf{x})}$$ you'll get that $$(-\omega^{2} + \mathbf{k}\cdot \mathbf{k})\tilde{A} = -M^{2}\tilde{A}$$, where $$\tilde{A}$$ is the Fourier transform of A. From this we get that $$-\omega^{2} + \mathbf{k}\cdot \mathbf{k} = -M^{2}$$ because the coefficients are all scalars and none are differential operators. This is a reformulation of the mass-energy-momentum relation.
So how can we get something to do with just E and not $$E^{2}$$? Through particular arguments (see Chapter 1 of 'Quantum Theory of Fields' by Weinberg) Dirac realised he needed a
first order operator to apply to a spinor. But the operator has to square to something which is the wave operator on
all components of the spinor. Clearly something with a second order time derivative and likewise spatial derivative needs to be of the form $$D = a\partial_{t} + \mathbf{b}\cdot \nabla$$. Except such a thing doesn't square to the wave operator no matter the scalar values of a and the components of b. But if you make them all matrices you can do it, then the operator you want at the end is the wave operator
times the identity matrix. Using the whole anticommutation relations thing I've explained multiple times you can deduce the conditions the matrices must satisfy, the Dirac algebra relations. In what follows I probably drop factors of i or -1 but that isn't relevant to what I'm getting at here.
Then what happens if you have $$D = \gamma^{0}\partial_{t} + \gamma^{i}\partial_{i}$$ is a
matrix operator, which when you apply to the spinor you get $$D \psi = m\psi = \left( \gamma^{0}\partial_{t} + \gamma^{i}\partial_{i} \right) \psi$$. If you Fourier transform this you get the usual alteration of the coefficients, $$m\tilde{\psi} \propto \left( \gamma^{0}\omega + \gamma^{i}k_{i} \right) \psi$$. However, unlike the scalar wave equation case you cannot say that the coefficients are equal, ie $$-\omega + \sum_{i}k_{i} + m = 0$$ because there's matrices involves and these matrices are not all the same. If you expanded out $$m\tilde{\psi} \propto \left( \gamma^{0}\omega + \gamma^{i}k_{i} \right)\tilde{\psi}$$ in terms of spinor components you'd find you
don't get $$-\omega + \sum_{i}k_{i} + m$$ in front of each term.
Since you no doubt haven't got a clue what I'm talking about since you don't know spinor matrix behaviour I'll use a
previous example. In 1+1 dimensions the Dirac operator could be written as $$D = \gamma^{0}\partial_{t} + \gamma^{1}\partial_{x} = \left( \begin{array} i\partial_{t} & \partial_{x} \\ \partial_{x} & -i\partial_{t} \end{array} \right)$$. This hits the 2 component spinor $$\left( \begin{array}{c} \psi_{1} \\ \psi_{2} \end{array} \right)$$ as $$D\psi = \left( \begin{array} i\partial_{t} & \partial_{x} \\ \partial_{x} & -i\partial_{t} \end{array} \right)\left( \begin{array}{c} \psi_{1} \\ \psi_{2} \end{array} \right) = m \psi = m\left( \begin{array}{c} \psi_{1} \\ \psi_{2} \end{array} \right)$$. So we Fourier transform each side and the equation reduces to $$\left( \begin{array} -\omega & ik \\ ik & \omega \end{array} \right)\left( \begin{array}{c} \tilde{\psi}_{1} \\ \tilde{\psi}_{2} \end{array} \right) = m\left( \begin{array}{c} \tilde{\psi}_{1} \\ \tilde{\psi}_{2} \end{array} \right)$$ which gives up a pair of equations, $$(-\omega-m) \tilde{\psi}_{1} + ik\tilde{\psi}_{2} = 0$$ and $$ik\tilde{\psi}_{1} + (\omega-m)\tilde{\psi}_{2} = 0$$. Clearly this hasn't reduced to $$\omega = m+k$$, there's mixing between the different components.
To see this in the proper light it's useful to consider $$D^{2}\psi$$. As I previously showed $$D^{2} = \left( \begin{array} -\partial_{t}^{2}+\partial_{x}^{2} & 0 \\ 0 & -\partial_{t}^{2}+\partial_{x}^{2} \end{array} \right) = \mathbb{I}(-\partial_{t}^{2}+\partial_{x}^{2})$$. This is diagonal, there is no mixing between the different components of whatever spinor I apply it to. This is the power of the Dirac algebra, in order to get a first order operator which squares to something that acts like the wave operator on each component
separately you have to use a
non-diagonal matrix operator. If it could be diagonal you'd not need matrices at all! But this non-diagonalness (which isn't really a word) is why you can't just say "Square root both sides of $$E^{2} = M^{2} + p^{2}$$ to get $$E = M+p$$!", because the squaring was done using matrices, so you can't unsquare the expression ignoring that fact.
As I said to you in the original thread, this is a
fundamental thing in the Dirac equation. The book of Weinberg I mentioned devotes the whole chapter to discussing how Dirac come to realise this requirement, the methods he considered, the algebra he worked through and the implications of the results and that's just the introduction, never mind the
detail. It's something every course on QED and the Dirac equation will highlight in the extreme. You claim you are knowledgeable in the Dirac equation, well enough to perhaps handle university material in it. This is
repeated evidence you're not. This isn't a typo, this isn't a little slip, this isn't a transcription error, this isn't an "I was heading out to get drunk and I just wrote it down" mistake, it's a
fundamental gaping hole in your supposed knowledge. It undermines any claim to be knowledgeable in this stuff because it means you've
never do any real working calculations with the Dirac equation else you'd know all about this. Spinor components and matrix actions on them are the bane of many a student in this stuff. I speak from personal experience of getting lost in indices many times.
Seriously, give it a rest. How many times do you need to have your ignorance exposed before you stop lying?