= Solution
There are exactly two tree <Feynman diagrams>: the incident scalar can be absorbed before the final scalar is emitted, or after it. The internal fermion <four-momenta> are respectively $r_s=p+k=q+l$ and $r_u=p-l=q-k$. A scalar exchange diagram would require an absent scalar self-interaction, so there is no additional tree channel.
Define the <S-matrix> convention $S_{fi}=\delta_{fi}+i(2\pi)^4\delta^{(4)}(p+k-q-l)T_{s's}$. The two tree <scattering amplitudes> are
$$
iT^{(s)}_{s's}=\bar u(q,s')(ig\gamma^5)S_F(p+k)(ig\gamma^5)u(p,s),
\qquad
iT^{(u)}_{s's}=\bar u(q,s')(ig\gamma^5)S_F(p-l)(ig\gamma^5)u(p,s).
$$
Using the <Dirac propagator> from the preceding <Feynman rules>,
$$
\boxed{T_{s's}=-g^2\bar u(q,s')\gamma^5\left[
\frac{M-i(\not p+\not k)}{(p+k)^2+M^2-i0}
+\frac{M-i(\not p-\not l)}{(p-l)^2+M^2-i0}
\right]\gamma^5u(p,s)}.
$$
The two terms add with the same relative sign: neither diagram exchanges identical external <fermions>. Their interference must be retained when squaring the complete <scattering amplitude>. In the phase convention above, $\gamma^5(M-i\not r)\gamma^5=-(M+i\not r)$, which gives an equivalent simplified numerator. An overall phase depends on the <S-matrix> convention and does not change a <relativistic scattering cross-section>.
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