Collid'em

In the lab frame (taking $c = 1$ ), the four-momentum vector $L^\mu$ may be written as $L^\mu = (E+m, \overrightarrow{p})$ , where $\overrightarrow{p}$ is the momentum of the moving proton. If we contract this four-vector we get

$L^\mu L_\mu = (E+m)^2 - \left |\overrightarrow{p} \right |^2 = E^2 - \left |\overrightarrow{p} \right |^2 + 2Em + m^2.$

But according to Einstein's energy equation, $E^2 - \left |\overrightarrow{p} \right |^2 = m^2$ . Therefore one may rewrite the above contraction as

$L^\mu L_\mu = 2m^2 + 2Em.$

Now, there exists a frame (called center of momentum frame) where the total momentum of the system is null. The existence of such frame is easy to prove in this case (which I encourage you to do), but it also exists for any other system. In this case, the four-momentum vector in the center of momentum frame $M^\mu$ after the collision (one may evaluate the four-momentum before or after the collision and it must be the same) is $M^\mu = (4m, \overrightarrow{0})$ since there are $4$ particles with mass $m$ . As the contraction of the four-momentum vector is a scalar, it must be the same, no matter what frame we're using:

$L^\mu L_\mu = M^\mu M_\mu \;\;\Rightarrow \;\; 2m^2 + 2Em = 16m^2$

$\therefore E = 7m.$