Consider LHS of option A:
¬∀x[ P(x) ∨ ∃y[ Q(y) ∧ P(y) ] ]
Consider RHS of option A:
∃x[ ¬P(x) ∧ ∀y[ (P(y) → ¬Q(y)) ∨ (Q(y) → ¬P(y)) ] ]
= ∃x[ ¬P(x) ∧ ∀y[ (¬P(y) ∨ ¬Q(y)) ∨ (¬Q(y) ∨ ¬P(y)) ] ]
= ∃x[ ¬P(x) ∧ ∀y[ (¬Q(y) ∨ ¬P(y)) ∨ (¬Q(y) ∨ ¬P(y)) ] ] (∵ A ∨ B = B ∨ A)
= ∃x[ ¬P(x) ∧ ∀y[ ¬Q(y) ∨ ¬P(y) ] ] (∵A ∨ A = A )
= ∃x[ ¬P(x) ∧ ¬[ ∃y[ Q(y) ∧ P(y) ] ] ] $[\neg(\forall xP(x))\equiv\neg \forall x\neg P(x)\equiv\exists x\neg P(x)]$
= ∃x[ ¬[ P(x) ∨ ∃y[ Q(y) ∧ P(y) ] ] ] $[\neg(\exists xP(x))\equiv\neg \exists x\neg P(x)\equiv\forall x\neg P(x)]$
= ¬∀x[ P(x) ∨ ∃y[ Q(y) ∧ P(y) ] ]
= LHS
So option (A) is the answer.