= Solution
Apply the distributive law of <Boolean algebra> to the two <logical conjunctions>. With $A=P$, $B=\neg Q$, $C=\neg P$, and $D=Q$, the identity $(A\land B)\lor(C\land D)\equiv(A\lor C)\land(A\lor D)\land(B\lor C)\land(B\lor D)$ gives
$$
\begin{aligned}
(P\land\neg Q)\lor(\neg P\land Q)
&\equiv(P\lor\neg P)\land(P\lor Q)\land(\neg Q\lor\neg P)\land(\neg Q\lor Q)\\
&\equiv\top\land(P\lor Q)\land(\neg P\lor\neg Q)\land\top\\
&\equiv\boxed{(P\lor Q)\land(\neg P\lor\neg Q).}
\end{aligned}
$$
The second step uses complementation and commutativity in <Boolean algebra>; the last uses the identity law for <logical conjunction>. The result is in <conjunctive normal form>: its first <Boolean clause> requires at least one input to be true, and its second <Boolean clause> prohibits both inputs being true. Together they express <exclusive or>.
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