Associativity applies only when the connectives involved are exclusively $\land$ or exclusively $\lor$:
$$p \land q \land r \equiv (p \land q)\land r \equiv p \land (q\land r)$$
$$p \lor q \lor r \equiv (p \lor q)\lor r \equiv p \lor (q\lor r)$$
Because of associativity of $\lor$ and $\land$, parentheses are not necessary to define expressions like those above.
Your statement, however:
$$p \land \lnot q \lor q \land \lnot r \lor \lnot p \lor r \tag{given}$$
has mixed connectives, and so associativity does not apply across all possible groupings.
Please note: as stated, your (given) expression is not well-defined without parentheses. That is, without parentheses, it is ambiguous; it can be read any number of ways, most of which are not equivalent. Does it mean connect from left to right?:
$$(((((p\land \lnot q) \lor q) \land) \lnot r)\lor\lnot p) \lor r\;?\tag{1}$$
Or does it mean this?
$(p \land \lnot q) \lor (q \land \lnot r) \lor (\lnot p \lor r)\;?\tag{2}$
or any number of other possible ways of grouping with parentheses?
In general, when you have an expression like $(2)$ above, you need to apply the Distributive Laws to distribute over another connective:
For example $$p \land (q \lor r) \equiv (p \land q) \lor (p \land r)$$
$$p \lor (q\land r) \equiv (p \lor q) \land (p\lor r)$$
With :
$(¬p∧¬r)∨(¬p∧q)∨(q∧¬r)∨(q∧q)$
you first apply $(q \land q) \equiv q$ to simplify it :
$(¬p∧¬r)∨(¬p∧q)∨(q∧¬r)∨q$.
Consider now : $(q∧¬r)∨q$; applying Absorption Laws : $p∨(p∧q) \equiv p$, we have that it is equivalent to $q$.
Thus :
$[(¬p∧¬r)∨(¬p∧q)]∨[(q∧¬r)∨q] \equiv (¬p∧¬r)∨(¬p∧q)∨q$.
Apply it again to : $(¬p∧q)∨q$, which is again equivalent to $q$, to get :
$(¬p∧¬r)∨[(¬p∧q)∨q] \equiv (¬p∧¬r)∨q$.
Regarding absorption :
$p∨(p∧q) \equiv p$
you can prove it with Natural Deduction or verify it via truth table :
if $p$ is TRUE, then $p \lor (p \land q)$ is TRUE, by truth table for $\lor$;
if $p$ is FALSE, then $(p \land q)$ is FALSE, by truth table for $\land$; so both $p$ and $(p \land q)$ are FALSE, and thus $p∨(p∧q)$ is FALSE, by truth table for $\lor$.
Best Answer
Intuitively
Notice that all the connectives are "or". In general, if you only have or's you can parenthesize the expression any way you want and it doesn't make a difference. This is because any way you parenthesize it, the expression will be true if at least one variable is true, and it will be false if every variable is false.
So I would just think of the expression as
$$¬p∨q ∨ ¬p∨r$$
And then switch the $q$ and $¬p$ (Commutative Law), since intuitively again it doesn't matter what order they're in it (for the same reason as above (consider when true/false)).
Thus we can transform it to
$$(¬p∨¬p) ∨ (q∨r)$$