Tracing the circuit: $Y=\overline{\overline{(A\cdot B)}\cdot (C+D)}$.

By De Morgan, $Y=(A\cdot B)+\overline{(C+D)}$.

Equivalently, $Y=1$ when either $A=B=1$, or both $C=0$ and $D=0$.

Row-by-row evaluation:

(1,1,0,1): $A\cdot B=1$, so $Y=1$.

(0,0,1,1): $A\cdot B=0$, $\overline{C+D}=0$, so $Y=0$.

(1,0,1,0): $A\cdot B=0$, $\overline{C+D}=0$, so $Y=0$.

(1,1,1,1): $A\cdot B=1$, so $Y=1$.

Outputs: 1, 0, 0, 1 — matches option (2).