= Solution
Cross the second incoming heavy particle and one outgoing heavy particle in <2/d/i> to obtain $\Phi\bar\Phi\to\Phi\bar\Phi$. Panel (e) of <1/b>'s figure shows the resulting <box diagram>. Its four light-charge arrows again alternate source and sink corners; it is the only connected order-$g^4$ topology. There is no tree exchange because every cubic <interaction vertex> has only one heavy leg. With $|\mathbf p|\ll M\ll m$, crossing does not affect the zero-external-<momentum> loop integral, so
$$
\boxed{\mathcal M_{\Phi\bar\Phi}=\frac{g^4}{96\pi^2m^4}
\left[1+O\left(\frac{M^2}{m^2},\frac{|\mathbf p|^2}{m^2}\right)\right].}
$$
An independent check is to integrate out the <complex scalar field> $\phi$ at constant $\Phi$. Its two real-component squared masses are $m^2\pm g|\Phi|$. The one-loop <quantum effective potential> is therefore
$$
V_1(\Phi)=\frac12\int\frac{d^4k_E}{(2\pi)^4}
\left[\log(k_E^2+m^2+g|\Phi|)+\log(k_E^2+m^2-g|\Phi|)\right].
$$
Expanding the logarithms to fourth order in $\Phi$ gives
$$
V_1\supset-\frac{g^4|\Phi|^4}{4}\int\frac{d^4k_E}{(2\pi)^4}\frac1{(k_E^2+m^2)^4}
=-\frac{g^4}{384\pi^2m^4}(\Phi^\dagger\Phi)^2.
$$
Thus the <effective Lagrangian> has the positive local coefficient $g^4/(384\pi^2m^4)$. Its $2!2!$ external attachments give the common <scattering amplitude> $g^4/(96\pi^2m^4)$ in both channels. This also checks the box multiplicity and sign.
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