For proving $iff$, letting $P_j$ denot "$p$ is prime," i.e., to prove:
$\exists j(Pj \land j\neq 2 \land j\in[1, 100]) \iff (\text{blah happens}),$
and given that you've established/proved the forward direction ($\implies$), if you want to prove the reverse implication
$(\text{blah happens}) \implies \exists j(Pj \land j\neq 2 \land j\in[1, 100])$
by proving its contrapositive,
then you would need to establish that $\lnot \exists j(Pj \land j\neq2 \land j\in[1,100]) \implies \lnot \text{(blah happens)},\tag{1}$
or equivalently
$\forall j (\lnot Pj \lor j = 2 \lor j\notin[1,100]) \implies \lnot \text{(blah happens)}.\tag{2}$
EDIT as requested by the OP:
To understand why $(1)$ and $(2)$ are equivalent, with the understanding that each represents the contrapositive of $(\text{blah happens}) \implies \exists j(Pj \land j\neq 2 \land j\in[1, 100]) $
We start from $(1)$:
$\lnot \exists j(Pj \land j\neq2 \land j\in[1,100]) \implies \lnot \text{(blah happens)}.\tag{1}$
We switch to the universal quantifier while moving the negation inward gives us the equivalent statement:
$\forall j[\lnot(Pj \land j\neq2 \land j\in[1,100])] \implies \lnot \text{(blah happens)}.\tag{1.a}$
Then by DeMorgan's law $(1.a)$ is equivalent to $(1.b)$:
$\forall j[(\lnot Pj \lor (\lnot (j\neq2) \lor (\lnot(j\in[1,100]))] \implies \lnot \text{(blah happens)},\tag{1.b}$
which you can see is equivalent to $(2)$:
$\forall j[(\lnot Pj \lor (j = 2) \lor (j\notin[1,100])] \implies \lnot \text{(blah happens)}\tag{2}$