= Solution
Evaluate the <two-body Lorentz-invariant phase space> in the decaying particle's rest frame, $p=(m,\mathbf0)$. The spatial delta function sets $\mathbf q_2=-\mathbf q_1$. Both massless particles have energy $q=|\mathbf q_1|$, leaving
$$
d\Pi_2=\frac1{16\pi^2}\,dq\,d\Omega\,\delta(m-2q)
=\frac{d\Omega}{32\pi^2}.
$$
Here the factor $1/2$ in the second equality is the Jacobian of the energy delta function. Integrating the solid angle gives $\int d\Pi_2=1/(8\pi)$. If the decay is isotropic, the squared amplitude is angle independent, so
$$
\boxed{\Gamma=\frac1{2m}\frac{|\mathcal A|^2}{8\pi}
=\frac{|\mathcal A|^2}{16\pi m}.}
$$
This uses distinguishable final particles or an amplitude/phase-space convention with any required identical-particle symmetry factor already included. Two genuinely identical final bosons require an additional $1/2!$ if the full angular domain is integrated without such a factor. The complex particle and antiparticle in the following parts are distinct, so no such division is required there.
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