= Solution
Work in the parent <rest frame>, $p=(M,\mathbf0)$. For massless daughters, the spatial <Dirac delta> fixes $\mathbf q_2=-\mathbf q_1$ and $E_1=E_2=q$. The <Lorentz invariance> two-body <Lorentz-invariant phase-space measure> becomes
$$
\begin{aligned}
d\Pi_2&=\frac{(2\pi)^4}{(2\pi)^6}\frac{q^2dq\,d\Omega}{4q^2}\,\delta(M-2q)\\
&=\frac{dq\,d\Omega}{16\pi^2}\,\delta(M-2q)
\quad\Longrightarrow\quad
\boxed{d\Pi_2=\frac{d\Omega}{32\pi^2}.}
\end{aligned}
$$
The factor $1/2$ in the last step is the Jacobian of the energy-conservation <Dirac delta>. Consequently the labelled two-body <decay width> is
$$
\boxed{\Gamma_{\mathrm{labelled}}=\frac1{64\pi^2M}\int d\Omega\,|\mathcal A|^2.}
$$
For two identical final <particles>, the full angular integral counts each unordered final state twice, so the physical <decay width> includes an additional factor $1/2!$. This factor is needed for both channels in part e; it is separate from any vertex combinatorics already included in the <Feynman rule>.
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