Answer
See the detailed answer below.
Work Step by Step
Since one of the initial protons was at rest and the other was moving which means that there is an initial momentum and energy; and since the 4 products' protons are having the same mass and the same speed, then we can consider them as one object which mass is given by $m=4m_{\rm p}$.
Now we need to find the $Q$-value of the reaction;
$$Q=2Q_{\rm p}-4Q_{\rm p}$$
$$Q=(2m_{\rm p}-4m_{\rm p})c^2=-2m_pc^2\tag 1$$
According to energy conservation;
$$E_{\rm p}+E_{\rm p,rest}=E_{\rm 4p}$$
Where $E_{\rm 4p}$ is the energy of the four product protons which is given by $E=KE+mc^2$;
$$\overbrace{E_{\rm p}}^{KE+mc^2}+\overbrace{E_{\rm p,rest}}^{\overbrace{KE}^{0}+mc^2}=KE_{\rm 4p}+4m_pc^2$$
$$KE_{\rm p,max}+m_pc^2 +m_pc^2 =KE_{\rm 4p}+4m_pc^2$$
Solving for $KE_{\rm p,max}$ which is the threshold energy
$$KE_{\rm p,max} =KE_{\rm 4p}+2m_pc^2\tag 2$$
According to momentum conservation;
$$p_{\rm p}+0=p_{\rm 4p}$$
Multiplying both sides by $c$ and then squaring both sides so we can use the formula of $E^2=p^2c^2+m^2c^4$ which gave us $p^2c^2=E^2-m^2c^4$
$$p_{\rm p}^2c^2=p_{\rm 4p}^2c^2$$
So that;
$$E_{\rm p}^2-m_{\rm p}^2c^4=E_{\rm 4p}^2-m_{\rm 4p}^2c^4$$
$$E_{\rm p}^2-m_{\rm p}^2c^4=E_{\rm 4p}^2-(4m_{\rm p})^2c^4$$
$$E_{\rm p}^2 =E_{\rm 4p}^2-15m_{\rm p}^2c^4$$
$$(KE_{\rm p,max}+m_pc^2 )^2 =(KE_{\rm 4p}+4m_pc^2)^2-15m_{\rm p}^2c^4$$
Plugging $KE_{\rm 4p}$ from (2); $KE_{\rm 4p}=KE_{\rm p,max} -2m_pc^2$;
Thus;
$$(KE_{\rm p,max}+m_pc^2 )^2 =(KE_{\rm p,max} -2m_pc^2+4m_pc^2)^2-15m_{\rm p}^2c^4$$
$$(KE_{\rm p,max}+m_pc^2 )^2 =(KE_{\rm p,max} +2m_pc^2)^2-15m_{\rm p}^2c^4$$
$$ KE^2_{\rm p,max}+2 m_pc^2KE_{\rm p,max}+m^2_pc^4 =\\
KE_{\rm p,max}^2+4m_pc^2KE_{\rm p,max} +4m^2_pc^4 -15m_{\rm p}^2c^4$$
$$ m^2_pc^4 -( 4m^2_pc^4 -15m_{\rm p}^2c^4) = 4m_pc^2KE_{\rm p,max} -2 m_pc^2KE_{\rm p,max}$$
Therefore;
$$ 2m_pc^2KE_{\rm p,max} =12m^2_pc^4 $$
$$ KE_{\rm p,max} =\dfrac{12m^2_pc^4 }{ 2m_pc^2}=6m_pc^2$$
Divide by (1);
$$ \dfrac{ KE_{\rm p,max} }{Q} =\dfrac{6m_pc^2}{2m_pc^2}=|-3|=3$$
And hence;
$$ \boxed{KE_{\rm p,max} =3Q}$$
