Using distributive law: ¬(p ∨ (¬p ∧ q)) =¬((p ∨¬p)∧ (p∨q)), equals ¬(1∧(p∨q))=¬(p∨q), using DeMorgan's Law
¬(p∨q)=¬p∧¬q , hence the equivalence is true
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