You can solve it "embedding" the induction into the formula itself.
You have that the "general case" can be written as :
$$[(P_0 \lor P_1 \lor P_2\lor\ldots\lor P_n) \lor P_{n+1}] \rightarrow Q $$
i.e. $[\mathcal{A} \lor P_{n+1}] \rightarrow Q $
But $P \rightarrow Q$ is $\lnot P \lor Q$, so that the above formula is :
$\lnot [\mathcal{A} \lor P_{n+1}] \lor Q $
Using De Morgan :
$[\lnot \mathcal{A} \land \lnot P_{n+1}] \lor Q $
now distribute :
$(\lnot \mathcal{A} \lor Q)\land (\lnot P_{n+1} \lor Q)$
i.e.
$(\lnot \mathcal{A} \lor Q) \land (P_{n+1} \rightarrow Q)$.
Now, what about $(\lnot \mathcal{A} \lor Q)$ ?
It is :
$\lnot (P_0 \lor P_1 \lor P_2\lor\ldots\lor P_n) \lor Q$
and using "general" De Morgan we have :
$(\lnot P_0 \land \lnot P_1 \land \lnot P_2 \land \ldots \land \lnot P_n) \lor Q$
Distribute again :
$(\lnot P_0 \lor Q) \land \ldots \land (\lnot P_n \lor Q)$
that is :
$(P_0 \rightarrow Q) \land \ldots \land (P_n \rightarrow Q)$
But this was $(\lnot \mathcal{A} \lor Q)$, so that the complete formula will be :
$(P_0 \rightarrow Q) \land \ldots \land (P_n \rightarrow Q) \land (P_{n+1} \rightarrow Q)$.