Recall that
$\begin{align}E^2 = p^2 c^2 + m^2 c^4 \label{eqn:1}\end{align}$

The two decay products must be back to back in order to conserve three momentum (the initial three momentum is $0$ ). Let $(E/c,0,0,p)$ be the four-vector of the massless particle. Equation (1) with $m$ replaced by $0$ implies that $E/c = p$ . So, the four-momentum of the massless particle is $(p,0,0,p)$ . By three-momentum conservation, the three-momentum of the massive particle must be $(0,0,-p)$ . From equation (1), the four-momentum must then be $(\sqrt{p^2 + m^2c^2},0,0,-p)$ . The initial energy is $M c^2$ , so by energy conservation,
$\begin{align*}Mc = \sqrt{p^2 + m^2c^2} + p\end{align*}$
Squaring both sides and solving for $p$ yields
$\begin{align*}p = \boxed{\frac{M^2 - m^2}{2M}}\end{align*}$
Therefore, answer (B) is correct.

Solution to 2008 Problem 40