= Solution
{c}
Label the <four-momenta> as $e^-(p_1)\mu^-(p_2)\to e^-(p_3)\mu^-(p_4)$. There is one <tree-level Feynman diagram>, the $t$-channel <photon> exchange shown in the lower-left panel. The two lepton species are distinct, so there is no exchanged-final-lepton diagram. With $s=(p_1+p_2)^2$, $t=(p_1-p_3)^2$ and $u=(p_1-p_4)^2$, the <Feynman rules> give
$$
i\mathcal M=\frac{ie^2}{t}[\bar u(p_3)\gamma^\mu u(p_1)][\bar u(p_4)\gamma_\mu u(p_2)].
$$
The massless <fermion spin sum> is $\sum_ru_r(p)\bar u_r(p)=\not p$. Averaging the two initial <spins> of each lepton supplies $1/4$, so
$$
\langle|\mathcal M|^2\rangle=\frac{e^4}{4t^2}\operatorname{tr}(\not p_3\gamma^\mu\not p_1\gamma^\nu)\operatorname{tr}(\not p_4\gamma_\mu\not p_2\gamma_\nu).
$$
The <gamma matrix trace identities> imply, for arbitrary null external <four-momenta>,
$$
\operatorname{tr}(\not a\gamma^\mu\not b\gamma^\nu)\operatorname{tr}(\not c\gamma_\mu\not d\gamma_\nu)=32[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)].
$$
To see the contraction, each trace is $4(a^\mu b^\nu+a^\nu b^\mu-g^{\mu\nu}a\cdot b)$; the terms proportional to $(a\cdot b)(c\cdot d)$ cancel in four dimensions, leaving the displayed two products. Here $p_1\cdot p_2=p_3\cdot p_4=s/2$ and $p_1\cdot p_4=p_2\cdot p_3=-u/2$. Therefore
$$
\boxed{\langle|\mathcal M|^2\rangle=2e^4\frac{s^2+u^2}{t^2}.}
$$
The requested proportionality constant is $2e^4=32\pi^2\alpha_{\mathrm{em}}^2$, where $\alpha_{\mathrm{em}}$ is the <fine-structure constant>. Both numerator terms are required by the <spin sum>; no final-state statistical factor enters a squared <scattering amplitude> for distinct leptons.
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