We'll first apply, to the right-hand side of your biconditional, the following identity:
$$P \Rightarrow Q \equiv \lnot P \lor Q\quad \tag{logically equivalent}$$
Each side of the above equivalence implies the other (they are logically equivalent).
Applying this to the right had side of where you left off $(1)$, we get $(2)$:
$$((P\wedge Q)\vee (\neg P \wedge \neg Q))\iff((P\Rightarrow Q)\wedge (Q\Rightarrow P))\tag{1}$$
$$((P\wedge Q)\vee (\neg P \wedge \neg Q))\iff(\lnot P \lor Q)\wedge (\lnot Q \lor P))\tag{2}$$
Try using distribution on the right hand side (twice) until you arrive at an equivalent expression as the left hand side of the biconditional. Putting this all together, we get:
$$\begin{align} ((P\land Q) \lor (\lnot P \land \lnot Q))
& \iff ((P \Rightarrow Q) \land (Q \Rightarrow P)) \\ \\
& \iff ((\lnot P \lor Q) \land (\lnot Q \lor P)) \\ \\
& \iff [\lnot P \land (\lnot Q \lor P)] \lor [Q \land (\lnot Q \lor P)] \\ \\
& \iff [(\lnot P \land \lnot Q) \lor (\lnot P \land P)] \lor [(Q\land \lnot Q) \lor (Q \land P)] \\ \\
& \iff (\lnot P \land \lnot Q) \lor F \lor F \lor (Q \land P) \\ \\
& \iff (\lnot P \land \lnot Q) \lor (Q\land P) \\ \\
& \iff ((P \land Q) \lor (\lnot P \land \lnot Q))
\end{align}$$
By using the given identity, applying the distributive law twice, and remembering that $A \land \lnot A \equiv F\;$ and $\;A \lor F \equiv A$, we have obtained, from the right hand side of our biconditional, the equivalent expression on the left-hand side of $(2)$.
$$(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$$
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":
$$\begin{align} (P \leftrightarrow Q) &\equiv (P \to Q) \wedge (Q \to P) \tag{correct}\\ \\
&\equiv (\color{blue}{\bf \lnot P \lor Q}) \land (\color{red}{\bf \lnot Q \lor P})\tag{correct} \\ \\
&\equiv \Big[\color{blue}{\bf \lnot P} {\land} \color{red}{\bf(\lnot Q \lor P)}\Big] \color{blue}{\lor} \Big[\color{blue}{\bf Q} \land \color{red}{\bf (\lnot Q \lor P)}\Big]\tag{DL}\\ \\
& \equiv \Big[(\color{blue}{\bf \lnot P} \land \color{red}{\bf\lnot Q)} \lor (\color{blue}{\bf\lnot P} \land \color{red}{\bf P})\Big] \lor \Big[(\color{blue}{\bf Q} \land \color{red}{\bf \lnot Q}) \lor (\color{blue}{\bf Q} \land \color{red}{\bf P})\Big]\tag{DL} \\ \\
&\equiv \Big[({\lnot P}\land \lnot Q) \lor \text{False}\Big] \lor \Big[\text{False} \lor (Q \land P)\Big]\tag{why?} \\ \\
&\equiv (P \land Q) \lor (\lnot P \land \lnot Q)\tag{why?}\end{align}$$
Best Answer
Umm... maybe I am missing something, but
if $p$ is false, $q$ is true and $r$ is false, then we have that
$(p\rightarrow q)\wedge(q\rightarrow r)$ is false
$p\rightarrow (q\wedge r)$ is true.
So I don't see how you can prove the equivalence.