The incoming photon is absorbed by the electron, producing a virtual electron with momentum $k = p + q$, which then emits the outgoing photon.
The computational chain
From diagram to number—every step of the amplitude calculation.
The diagram
The incoming photon is absorbed at vertex 1, creating a virtual electron with momentum $k = p + q$ and $k^2 = s$. The virtual electron then emits the outgoing photon at vertex 2.
Feynman rules expression
$$i\mathcal{M}_s = (-ie)^2\,\epsilon_\mu(q)\,\epsilon_\nu^*(q')\,\bar{u}(p')\gamma^\nu\frac{i(\not{p}+\not{q}+m)}{(p+q)^2-m^2}\gamma^\mu u(p)$$
Unlike Moller or Bhabha, this is a single fermion line: the electron enters, absorbs a photon, propagates, emits a photon, and exits. The two $\gamma$ matrices are sandwiched between the same pair of spinors. The external photons contribute polarization vectors $\epsilon_\mu$ and $\epsilon_\nu^*$.
Square and sum over spins and polarizations
$$\overline{|\mathcal{M}_s|^2} = \frac{1}{4}\frac{e^4}{(s-m^2)^2}\,\mathrm{Tr}\left[(\not{p}'+m)\gamma^\nu(\not{p}+\not{q}+m)\gamma^\mu(\not{p}+m)\gamma_\mu'(\not{p}+\not{q}+m)\gamma_\nu'\right]\cdot(-g^{\mu\mu'})(-g^{\nu\nu'})$$
Average over initial electron spin (1/2) and initial photon polarization (1/2). Sum over final spins and polarizations. The photon polarization sum $\sum_\lambda \epsilon_\mu^\lambda\epsilon_\nu^{\lambda*} \to -g_{\mu\nu}$ replaces each polarization vector pair with a metric tensor. This replacement sums over all four polarization states (including unphysical ones), but the Ward identity guarantees the unphysical polarizations contribute zero for gauge-invariant amplitudes.
Structure of the trace
$$\overline{|\mathcal{M}_s|^2} = \frac{e^4}{4(s-m^2)^2}\,\mathrm{Tr}\left[(\not{p}'+m)\gamma^\nu(\not{p}+\not{q}+m)\gamma^\mu(\not{p}+m)\gamma_\mu(\not{p}+\not{q}+m)\gamma_\nu\right]$$
After contracting the polarization metrics, this is a single trace of eight gamma matrices (or fewer after using contraction identities). The internal propagator numerator $(\not{p}+\not{q}+m)$ appears twice because it comes from $|\mathcal{M}|^2 = \mathcal{M}\mathcal{M}^*$. Apply $\gamma^\mu\gamma_\alpha\gamma_\mu = -2\gamma_\alpha$ to simplify the inner contractions.
Simplify using contraction identities
$$\gamma^\mu(\not{p}+m)\gamma_\mu = -2\not{p} + 4m$$
The contraction identity $\gamma^\mu\gamma^\alpha\gamma_\mu = -2\gamma^\alpha$ and $\gamma^\mu m\gamma_\mu = 4m$ reduce the eight-gamma trace to a six-gamma trace, then further. Unlike the Moller/Bhabha case (which factored into two separate traces), Compton produces a single long trace because both vertices sit on the same fermion line.
Result (full, with mass)
$$\overline{|\mathcal{M}_s|^2} \supset -2e^4\frac{m^2 - u}{m^2 - s}$$
The s-channel contribution to the full (massive) Compton amplitude. The mass terms cannot be dropped — the propagator denominator $m^2 - s$ vanishes in the massless limit, so the mass regulates the amplitude.