$\sim [(- p) \land q]$ is logically equivalent to

p \lor (\sim q)

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p \land (\sim q)

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\sim [p \land (\sim q)]

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\sim (p \lor q)

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Solution

The correct option is B $p \lor (\sim q)$
We know: $\sim (p \land q) \equiv\sim p \lor \sim q$
Hence $\sim [(- p) \land q] \equiv\sim (\sim p) \lor (\sim q) \equiv p \lor (\sim q)$ Using: $\sim (\sim p) = p$

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