Let the upper NAND output be $X=\overline{\bar{A}\cdot B}$ and the lower NAND output be $Y=\overline{A\cdot B}$.

The final AND gate yields $C=X\cdot Y=\overline{\bar{A}\cdot B}\cdot \overline{A\cdot B}$.

Applying De Morgan's theorem: $C=\overline{(\bar{A}\cdot B)+(A\cdot B)}=\overline{B(A+\bar{A})}=\overline{B}$.

So $C=\bar{B}$. Tabulating: when $B=0$, $C=1$; when $B=1$, $C=0$, independent of $A$. This matches option (2).