Find the simplest proposition logically equivalent to $(p \Rightarrow q) \Rightarrow ( p \Leftrightarrow q)$ Here is what I've gotten so far ... 
$$ (p\Rightarrow q)  \Rightarrow ( p \Leftrightarrow q)$$
$$\Leftrightarrow (\lnot p \vee q) \Rightarrow ((p\Rightarrow q) \wedge (q\Rightarrow p))$$
$$\Leftrightarrow(p \wedge \lnot q) \vee (\lnot p \vee q) \wedge (\lnot q \vee p)$$
this where I ended up, and I'm not sure if its even correct. If it is I'm not sure how to further simplify beyond this point. My best hunch is using associative law to group them differently and maybe simplify from there?
 A: Make sure you don't drop those parenteheses!
That is, your last step should result in:
$(p \land \neg q) \lor \color{red}((\neg p \lor q) \land (\neg q \lor p)\color{red})$
Now, from there, you can do a Distribution:
$((p \land \neg q) \lor \neg p \lor q) \land ((p \land \neg q) \lor \neg q \lor p)$
And then two more Distributions gets you:
$(p \lor \neg p \lor q) \land (\neg q \lor \neg p \lor q) \land (p \lor \neg q \lor p) \land (\neg q \lor \neg q \lor p)$
... and now you can really simplify things! ... but I'll leave that up to you.
Finally, let me show you a short-cut to do this:
TWo very handy equivalence principles are:
Reduction
$\neg p \land (p \lor q) = \neg p \land q$
$\neg p \lor (p \land q) = \neg p \lor 
q$
(In both forms, the $\neg p$ term 'reduces' the other term by removing the $p$)
Absorption
$p \lor (p \land q)$
$p \land (p \lor q)$
(In both forms, the $p$ term 'absorbs' the other term that includes a $p$) 
Applied to your statement:
$(p \to q) \to (p \leftrightarrow q)=$
$\neg (p \to q) \lor ((p \to q) \land (q \to p))\overset{Reduction!}{=}$
$\neg (p \to q) \lor (q \to p)=$
$(p \land \neg q) \lor \neg q \lor p\overset{Absorption!}{=}$
$\neg q \lor p=$
$q \to p$
