= Solution
\b[The printed integration labels need correction.] The parent's <four-momentum> is fixed, so the two <Lorentz-invariant phase-space measures> must integrate $p_2$ and $p_3$, rather than the printed $p_1$ and $p_2$. Write $P=p_1=(M,\mathbf0)$ and $p=p_2$, $k=p_3$. The physical <decay width> is
$$
\Gamma=\frac{\overline{|\mathcal M|^2}}{2M}
\int\frac{d^3p}{(2\pi)^3\,2E_p}
\frac{d^3k}{(2\pi)^3\,2E_k}
(2\pi)^4\delta^{(4)}(P-p-k).
$$
The spatial <Dirac delta function> sets $\mathbf k=-\mathbf p$. Performing the angular integral in <spherical coordinates>, with $E=\sqrt{p^2+m^2}$, gives the integrated <two-body decay phase space>
$$
\Phi_2=\frac1{4\pi}\int_0^\infty\frac{p^2\,dp}{E^2}\delta(M-2E).
$$
The positive root is $p_*=\sqrt{M^2/4-m^2}$. Since $d(2E)/dp=2p/E$, the <Dirac delta function> integration yields $\Phi_2=p_*/(4\pi M)$. Thus
$$
\boxed{\Gamma=\frac{\overline{|\mathcal M|^2}\,p_*}{8\pi M^2}
=\frac{g^2(M+2m)}{16\pi}\sqrt{1-\frac{4m^2}{M^2}},\qquad M>2m.}
$$
The two final particles have different species, so there is no <identical final-state symmetry factor>. The result tends to $g^2M/(16\pi)$ as $m\to0$ and vanishes at the two-particle <decay threshold>.
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