proving logical equivalence $(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$

$$(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$$

I'll start with your initial work, but instead of employing DeMorgan's as you did, we'll use the Distributive Law (DL), in two "steps":

$$\begin{align} (P \leftrightarrow Q) &\equiv (P \to Q) \wedge (Q \to P) \tag{correct}\\ \\ &\equiv (\color{blue}{\bf \lnot P \lor Q}) \land (\color{red}{\bf \lnot Q \lor P})\tag{correct} \\ \\ &\equiv \Big[\color{blue}{\bf \lnot P} {\land} \color{red}{\bf(\lnot Q \lor P)}\Big] \color{blue}{\lor} \Big[\color{blue}{\bf Q} \land \color{red}{\bf (\lnot Q \lor P)}\Big]\tag{DL}\\ \\ & \equiv \Big[(\color{blue}{\bf \lnot P} \land \color{red}{\bf\lnot Q)} \lor (\color{blue}{\bf\lnot P} \land \color{red}{\bf P})\Big] \lor \Big[(\color{blue}{\bf Q} \land \color{red}{\bf \lnot Q}) \lor (\color{blue}{\bf Q} \land \color{red}{\bf P})\Big]\tag{DL} \\ \\ &\equiv \Big[({\lnot P}\land \lnot Q) \lor \text{False}\Big] \lor \Big[\text{False} \lor (Q \land P)\Big]\tag{why?} \\ \\ &\equiv (P \land Q) \lor (\lnot P \land \lnot Q)\tag{why?}\end{align}$$

That's a great book. You'll learn a lot from it.

Justify the following: $$\begin{align} (\neg P \lor Q) \land (\neg Q \lor P) &\equiv[(\neg P\lor Q) \land \neg Q] \lor [(\neg P\lor Q)\land P] \\ &\equiv [(\neg P \land \neg Q) \lor (Q\land \neg Q)]\lor [(\neg P\land P)\lor (Q\land P)] \\ &\equiv (\neg P\land \neg Q) \lor (Q\land P).\end{align}$$