Solution

As in many other problems in this section, the author considers mass to be a measure of total energy. Then, let the center-of-mass system of two particles
$$E_1+E_2=Mc^2 \qquad \vec{p_1}+\vec{p_2}=0 \quad$$
$Note \ 1$: It follows from this that the sum of the projections of the momenta on any axis is equal to 0

Now, using the Lorentz transformations, we move to a system moving with velocity $v$ along the $x$ axis. For each particle, we can write:
\begin{pmatrix} E'/c \\ p'_x \\ p'_y \\ p'_z \end{pmatrix}= \begin{pmatrix} E/c\\ p_x \\ p_y \\ p_z \end{pmatrix} \cdot \begin{pmatrix} \gamma& \beta\gamma & 0 & 0 \\ \beta\gamma &\gamma& 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}= \begin{pmatrix} \frac{\gamma}{c}(E+vp_x)\\ \gamma(p_x+\beta\frac{E}{c}) \\ p_y \\ p_z \end{pmatrix}\tag{2}

Summing, we obtain:
$$E'=E'_1-E'_2=\gamma(E_1+vp_{x1}+E_2+vp_{x1})\tag{3}$$
$$\vec{P'}=\hat{x}(p'_{x1}+p'_{x2})+\hat{y}(p'_{y1}+p'_{y2})+\hat{z}(p'_{z1}+p'_{z2})$$
$$\vec{P'}=\gamma\hat{x}\left(p_{x1}+\beta\frac{E_1}{c}+p_{x2}+\beta\frac{E_2}{c}\right)+\hat{y}(p_{y1}+p_{y2})+\hat{z}(p_{z1}+p_{z2})\tag{4}$$
Concidering $Note \ 1$,
$$E'=M'c^2=\gamma(E_1+E_2+vp_{x1})=\gamma Mc^2 \quad so\quad M'=\gamma M \tag{5}$$
$$\vec{P'}=\gamma\hat{x}\left(\beta\frac{E_1+E_2}{c}\right)=\gamma \vec{v} \frac{Mc^2}{c^2}=\gamma M\vec{v} \tag{6}$$

Which is what we needed.