= Solution
For an allowed <decay channel> in <2/c/iii>, $|\mathcal M|^2=h^2$. Integrating the spatial <Dirac delta distribution> in the printed <Lorentz-invariant phase-space measure> sets $\mathbf p_2=-\mathbf p_1$. With $p=|\mathbf p_1|$ and $E_p=\sqrt{p^2+m^2}$, its remaining value is
$$
\begin{aligned}
I_2&=\int d\Phi_2=\frac1{4(2\pi)^2}\int\frac{d^3\mathbf p}{E_p^2}\,\delta(2E_p-M)\\
&=\frac1{4\pi}\int_0^\infty\frac{p^2\,dp}{E_p^2}\,\delta(2E_p-M).
\end{aligned}
$$
The positive root is $p_*=(M/2)\beta$, where $\beta=\sqrt{1-4m^2/M^2}$, and the <Dirac delta distribution> contributes the <Jacobian determinant> $E_*/(2p_*)$. Hence
$$
I_2=\frac{p_*}{8\pi E_*}=\frac{\beta}{8\pi},\qquad E_*=\frac M2.
$$
It follows that the \b[literal printed integral] evaluates to
$$
\boxed{\Gamma_{\rm printed}=\frac{h^2}{16\pi M}\sqrt{1-\frac{4m^2}{M^2}}.}
$$
The printed integration ranges count each final configuration twice, once as $(\mathbf p_1,\mathbf p_2)$ and once in the opposite order. For two identical final <bosons>, the physical <decay width> therefore requires an additional <identical final-state symmetry factor> $1/2!$:
$$
\boxed{\Gamma_{\rm phys}=\frac{h^2}{32\pi M}\sqrt{1-\frac{4m^2}{M^2}}\simeq\frac{h^2}{32\pi M}\quad(M\gg m).}
$$
There is one allowed <decay channel> per heavy <particle>, not an extra factor of two from summing the two charge-conjugate heavy species. Both expressions vanish below the <two-body decay threshold> $M=2m$, and the limiting value at threshold is zero. For the literally forbidden labels $\Phi\to\phi\phi$, the <decay width> is instead zero because its <decay amplitude> vanishes, as explained in <2/c/iii>.
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