= Solution
Use the <Feynman vertex> $-i\lambda$ for the scalar <Yukawa coupling>, the internal <scalar propagator> $i/(k^2-\mu^2+i0)$, incoming factors $u_r(p)$ and outgoing <Dirac adjoints> $\bar u_{r'}(p')$. Define $i\mathcal M$ as the coefficient in the connected <S-matrix> element after removing its overall momentum-conserving <Dirac delta function>. A direct pairing contributes
$$
i\mathcal M_t=(-i\lambda)^2\frac{i}{(p-p')^2-\mu^2+i0}\,[\bar u_{r'}(p')u_r(p)]\,[\bar u_{s'}(q')u_s(q)].
$$
The crossed pairing interchanges the identical outgoing <fermions> and therefore has a relative <fermionic sign> of minus. With $t=(p-p')^2$ and $u=(p-q')^2$, the result is
$$
\boxed{\mathcal M=-\lambda^2\left[\frac{[\bar u_{r'}(p')u_r(p)][\bar u_{s'}(q')u_s(q)]}{t-\mu^2+i0}
-\frac{[\bar u_{s'}(q')u_r(p)][\bar u_{r'}(p')u_s(q)]}{u-\mu^2+i0}\right].}
$$
It changes sign under exchange of the two outgoing momenta and spin labels, as required for identical <fermions>. No <fermion spin sum> or identical-particle phase-space factor belongs in this amplitude; those enter a <relativistic scattering cross-section> calculation.
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