= Solution
<Complex conjugation> interchanges the <Dolbeault operator> and <conjugate Dolbeault operator>. Thus $\partial\alpha=0$ implies $\bar\partial\bar\alpha=0$, and $\bar\alpha$ has type $(q,p)$. Its antiholomorphic degree is $p\geq1$, so the allowed <Dolbeault-Poincaré lemma> on the <polydisc> gives a form $\gamma\in\mathcal A^{q,p-1}$ satisfying $\bar\partial\gamma=\bar\alpha$.
Conjugate this identity and put $\beta=\bar\gamma$. This proves the <conjugate Dolbeault-Poincaré lemma>:
$$
\boxed{\beta\in\mathcal A^{p-1,q}(\mathcal P^n),\qquad\partial\beta=\alpha.}
$$
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