= Solution
A pure-type $d$-closed form is both $\partial$-closed and $\bar\partial$-closed, because these two derivatives have different types. Let $k=p+q$. The smooth <Poincaré lemma> on the <polydisc> gives $\eta\in\mathcal A^{k-1}_{\mathbb C}$ with $d\eta=\alpha$. We will modify $\eta$ by exact terms without changing $d\eta$.
Write $\eta=\sum_r\eta^{r,k-1-r}$. Starting with the smallest holomorphic degree, eliminate every component with $r<p-1$. Once the lower components are zero, the $(r,k-r)$ component of $d\eta$ says $\bar\partial\eta^{r,k-1-r}=0$, since $r<p$. Its antiholomorphic degree is positive, so the <Dolbeault-Poincaré lemma> supplies $\nu^{r,k-2-r}$ with $\bar\partial\nu=\eta^{r,k-1-r}$. Replace $\eta$ by $\eta-d\nu$: this removes the component at $r$ and changes only the next holomorphic degree.
Next, work downwards from the largest holomorphic degree and eliminate the components with $r>p$. Once the higher components are zero, the corresponding component of $d\eta$ gives $\partial\eta^{r,k-1-r}=0$. By part (b), write this as $\partial\nu^{r-1,k-1-r}$ and again subtract $d\nu$. This affects only the next lower holomorphic degree. The two finite procedures leave
$$
\eta=u+v,\qquad u\in\mathcal A^{p-1,q},\quad v\in\mathcal A^{p,q-1}.
$$
The components of $d\eta=\alpha$ outside type $(p,q)$ now give $\bar\partial u=0$ and $\partial v=0$, while its $(p,q)$ component gives $\alpha=\partial u+\bar\partial v$.
Since $q\geq1$, the <Dolbeault-Poincaré lemma> gives $u=\bar\partial a$ with $a$ of type $(p-1,q-1)$. Since $p\geq1$, part (b) gives $v=\partial b$ of the same type. The anticommutation identity therefore yields
$$
\alpha=\partial\bar\partial a+\bar\partial\partial b=\partial\bar\partial(a-b).
$$
This proves the <Bott-Chern Poincaré lemma> using precisely the permitted primitives:
$$
\boxed{H_{BC}^{p,q}(\mathcal P^n)=0\qquad(p,q\geq1).}
$$
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