= Solution
Boost along $+x$ at $0.6c$ into the <centre-of-momentum frame>. The composite is stationary. The incoming particle has $p'_x=1.25(2.4-0.6\times3)mc=0.75mc$, while the target has $p'_x=-0.75mc$. After decay the two <photons> have energies $Mc^2/2=1.6mc^2$ and opposite momenta.
Equal laboratory <photon> energies require the rest-frame <photons> to have zero <momentum> component along the boost: $E_{\rm lab}=\gamma(E'+Vp'_x)$ must be equal for two opposite <photons>. Their directions are therefore \b[$+y$ and $-y$ in the centre-of-momentum diagram], perpendicular to the incoming collision axis. Boosting those transverse <photons> gives exactly the laboratory momenta in part (a). A stationary composite has no direction-of-travel arrow in this frame.
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