Hint 2

At threshold all the final particles are at rest in the c.m. frame. Each initial proton must have half of the total rest energy of the final particles:

$$E_p=\left(M_p+{m_\pi\over 2}\right)c^2=1.006\ \hbox{\rm GeV}$$

In the lab frame, where proton 2 is at rest, use the fact that the total 4-momentum is conserved,

$${\sf p}_1+{\sf p}_2={\sf p}_3+{\sf p}_4+{\sf p}_\pi$$

The square of the total 4-momentum is an invariant. It can be calculated most easily in the c.m. frame,

$$({\sf p}_1+{\sf p_2})^2=({\sf p}_3+{\sf p}_4+{\sf p}_\pi)^2 =-(2M_p+m_\pi)^2c^2$$

Expanding the square of ${\sf p}_1+{\sf p}_2$ and using ${\sf p}^2=-M_2c^2$gives

$$2M_p^2c^2+2M_pE_1=(2M_p+m_\pi)^2c^2$$

This can now be solved to find the incoming proton energy

$$E_1=\left(M_p+2m_\pi+{m_\pi^2\over 2M_p}\right)c^2=1.218\ \hbox{\rm GeV}$$

The corresponding momentum of the proton is

$$p_1=\sqrt{{E_1^2\over c^2}-M_p^2c^2}=0.777\ \hbox{\rm GeV}$$