= Solution
At first order the <Dyson series>[Dyson expansion] inserts $iS_{\mathrm{int}}=\lambda\int d^4x\,\bar\psi_\alpha(x)(\gamma_5)_{\alpha\beta}\psi_\beta(x)\phi(x)$. Contract the <scalar field> at the <vertex> with the external <Scalar field>[scalar]. The four remaining <fermion> fields have two admissible pairings: both external <fermions> attach to the <vertex>, or the two <vertex> <fermions> contract with each other while the external pair contracts freely. <Wick's theorem> gives
$$
\boxed{\begin{aligned}
\langle T\phi(y)\psi_\rho(z)\bar\psi_\sigma(w)\rangle
={}&\lambda(\gamma_5)_{\alpha\beta}\int d^4x\,i\Delta_F(y-x)\Bigl[
iS_F(x-w)_{\beta\sigma}\,iS_F(z-x)_{\rho\alpha}\\
&\hspace{40pt}-iS_F(0)_{\beta\alpha}\,iS_F(z-w)_{\rho\sigma}\Bigr]+O(\lambda^3).
\end{aligned}}
$$
The minus sign in the second term is the <Fermionic sign>[fermion sign] of the closed <Wick contraction>[contraction]: $\langle T\bar\psi_\alpha(x)\psi_\beta(x)\rangle=-iS_F(0)_{\beta\alpha}$. The first term is connected; the second is a scalar-attached <fermion> <Tadpole diagram>[tadpole] multiplied by a disconnected free external <Dirac propagator>[fermion propagator]. The full time-ordered expectation includes both topologies, whereas the connected <generating functional> retains only the first. There is no order-$\lambda^2$ contribution in this pure Yukawa expansion because the total number of <scalar fields> would be odd.
For a <Parity>[parity-preserving] free propagator, the <Tadpole diagram>[tadpole] coefficient is $\operatorname{tr}[\gamma_5S_F(0)]$, whose integrand contains only $\operatorname{tr}\gamma_5$ and $\operatorname{tr}(\gamma_5\gamma^\mu)$. Both vanish. Thus the second term is a legitimate formal <Wick contraction> but is zero with a compatible regulator; <normal ordering> the <Interaction term>[interaction] removes it from the outset.
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