Use a truth table to verify this
¬(p ∧ q) ≡ ¬p ∨ ¬q.
Since (p ∧ q) ←→ ¬p ∨ ¬q is T in all cases, therefore
(p ∧ q) ≡ ¬p ∨ ¬q.
You could stop one step earlier by noticing that since the
columns for ¬(p ∧ q) and ¬p ∨ ¬q are identical, therefore
they’re logically equivalent.
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