I want to show that $$(¬P\wedge¬Q)\vee(¬P\wedge Q)\equiv¬P\wedge(¬Q\vee Q)$$ by one of the two Distributivity Laws: $$P\wedge(Q\vee R)\equiv(P\wedge Q)\vee(P\wedge R)$$ $$P\vee(Q\wedge R)\equiv(P\vee Q)\wedge(P\vee R).$$
I have tried to use both laws, without success. However, I managed to show that the statement holds using truth tables.
I think I may be missing something very obvious and/or simple, and I would appreciate any help.
Use the first one. The first one says that $P\land (Q\lor R)\equiv (P\land Q)\lor (P\land R)$. In the context of what you are trying to show, switch $P$ to $\neg P$, $Q$ to $\neg Q$, and $R$ to $Q$. Doing so gives the following: $$ \underbrace{P\land (Q\lor R)\equiv (P\land Q)\lor (P\land R)}_{\text{first distributive law before switch}}\overset{\text{switch}}{\equiv}\underbrace{(\neg P\land\neg Q)\lor(\neg P\land Q)\equiv\neg P\land(\neg Q\lor Q)}_{\text{first distributive law after switch}}. $$