= Solution
Write $p_x=\mathbb P(X=x)$, $\pi_x=\mathbb E(A\mid X=x)$, and $\mu_x=\mathbb E(Y\mid X=x)$. The functional is
$$
\beta=\mathbb E(AY)-\sum_xp_x\pi_x\mu_x.
$$
The first term has <influence function> $AY-\mathbb E(AY)$. Applying the product rule to the second term, including perturbations of $p_x$, $\pi_x$, and $\mu_x$, gives
$$
\pi(X)\mu(X)-\mathbb E[\pi(X)\mu(X)]
+\mu(X)\{A-\pi(X)\}
+\pi(X)\{Y-\mu(X)\}.
$$
Subtracting and using $\beta=\mathbb E(AY)-\mathbb E[\pi\mu]$ yields
$$
\psi(X,A,Y)
=\{A-\pi(X)\}\{Y-\mu(X)\}-\beta.
$$
Its expectation is zero because the first product has expectation $\mathbb E\{\operatorname{Cov}(A,Y\mid X)\}=\beta$.
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