Logic Gates question from JEE Main 2020 (2 Sep, Shift 2)

Answer: (d)

1. NAND: $P = \overline{AB}$. OR: $A + B$. AND: $P(A + B) = A\bar{B} + \bar{A}B$.
2. $Z = \overline{P + (A\bar{B} + \bar{A}B)}$. If $P = 1$, $Z = 0$.
3. $P = 1$ for $(1,0), (0,0), (0,1)$, so Z = 0 there.
4. For $(1,1)$: $P = 0$ and XOR $= 0$, so $Z = 1$.
5. Output: $0, 0, 1, 0$. So the answer is (d).