Match List-I with List-II: List-IList-II (a) p→q (i) ¬(q→¬p) (b) p∨q (ii) p∧¬q (c) p∧q (iii) ¬p→q (d)…

Using $p\to q\equiv\lnot p\lor q$: (a) $p\to q$ matches (iv) $\lnot p\lor q$; (b) $p\lor q\equiv\lnot p\to q$, matching (iii); (c) $\lnot(q\to\lnot p)\equiv\lnot(\lnot q\lor\lnot p)\equiv p\land q$, matching (i); and (d) $\lnot(p\to q)\equiv p\land\lnot q$, matching (ii). Pairing $p\lor q$ with $\lnot(q\to\lnot p)$ is wrong because that expression reduces to a conjunction, not a disjunction.