Using equivalence properties show that $[(p\land (\neg q)) \land (p\land (\neg q)) \lor (q\land (\neg r))] \iff (p \land (\neg q))$ I almost have the answer but I can't seem to figure out how to make the last contradiction disappear.
$[(p\land (\neg q)) \land (p\land (\neg q)) \lor (q\land (\neg r))] \iff (p \land (\neg q))$

(i) Commutativity and Associativity
$[(p\land (\neg q)) \land (p\land (\neg q)) \lor (q\land (\neg r))] \iff (\neg q) \land (p \land p) \land ((\neg q)\lor q) \land (\neg r)$
(ii) Idempotence and Contradiction 
$[(p\land (\neg q)) \land (p\land (\neg q)) \lor (q\land (\neg r))] \iff (\neg q) \land p \land$ 0 $\land (\neg r)$
(iii) Commutativity and De Morgan's Laws
$[(p\land (\neg q)) \land (p\land (\neg q)) \lor (q\land (\neg r))] \iff (p \land (\neg q)) \land (\neg((\neg$ 0 $)\lor r))$
(iv) Tautology
$[(p\land (\neg q)) \land (p\land (\neg q)) \lor (q\land (\neg r))] \iff (p \land (\neg q)) \land (\neg($ 1 $\lor r))$
(v) Tautology
$[(p\land (\neg q)) \land (p\land (\neg q)) \lor (q\land (\neg r))] \iff (p \land (\neg q)) \land (\neg($ 1 $))$
(vi) Contradiction
$[(p\land (\neg q)) \land (p\land (\neg q)) \lor (q\land (\neg r))] \iff (p \land (\neg q)) \land$ 0
Note: 1 means tautology and 0 means contradiction
 A: I don’t follow your first step; it certainly doesn’t follow just from associativity and commutativity. Your $\big((\neg q)\lor q\big)$ suggests that you’ve tried to use distributivity, but in that case you should also have $(\neg q)\lor(\neg r)$, among other things. In any case, the expression on the right-hand side after your step (i) is true only when $r$ is false, which is not the case for the target expression $p\land(\neg q)$, so you can’t possibly reduce yours to the desired target.
You have
$$\big(p\land(\neg q)\big)\land\big(p\land(\neg q)\big)\lor\big(q\land(\neg r)\big)\;.$$
If $\varphi\equiv p\land(\neg q)$ and $\psi\equiv q\land(\neg r)$, this is $\varphi\land\varphi\lor\psi$. Unfortunately, this is ambiguous unless you have a convention giving one of the connectives precedence over the other. If it means $(\varphi\land\varphi)\lor\psi$, then it reduces to $\varphi\lor\psi$, which is $\big(p\land(\neg q)\big)\lor\big(q\land(\neg r)\big)$, which you can then expand using distributivity to
$$\Big(\big(p\land(\neg q)\big)\lor q\Big)\land\Big(\big(p\land(\neg q)\big)\lor(\neg r)\Big)\;,$$
thence to
$$(p\lor q)\land\big((\neg q)\lor q\big)\land\big(p\lor(\neg r)\big)\land\big((\neg q)\lor(\neg r)\big)\;,$$
and, after reducing $(\neg q)\lor q$ to $1$, to 
$$(p\lor q)\land\big(p\lor(\neg r)\big)\land\big((\neg q)\lor(\neg r)\big)\;.\tag{1}$$
It’s not hard to check that if $r$ is false, $q$ is true, and $p$ is false, then $(1)$ is true but $p\land(\neg q)$ is false, so the two can’t be equivalent.
If $\varphi\land\varphi\lor\psi$ means $\varphi\land(\varphi\lor\psi)$, on the other hand, you can use an absorption law to reduce it to $\varphi$, which is exactly what you want. I suspect, therefore, that you’re missing a parenthesis, and that the left-hand side should be
$$\big(p\land(\neg q)\big)\land\Big(\big(p\land(\neg q)\big)\lor\big(q\land(\neg r)\big)\Big)\;.$$
