Truth table for system of four NAND gates as shown in figure is :

As shown in the logic diagram,

Output $Y=\overline{\overline{(A.\overline{A.B}})\overline{(B.\overline{A.B)}}}$.....eq(1)

When $A=0,B=0$ putting $A$ and $B$ in eq (1)$⟹$ $Y=0$

When $A=0,B=1$ putting $A$ and $B$ in eq (1) $⟹$ $Y=1$

When $A=1,B=0$ putting $A$ and $B$ in eq (1) $⟹$ $Y=1$

When $A=1,B=1$ putting $A$ and $B$ in eq (1) $⟹$ $Y=0$

Hence option $A$  is correct.