Which of the following circuits satisfies the $\operatorname{logic}$ condition $A=1, B=1$ and $D=1$ ?

In option (a)
where, $D=A \cdot \bar{B}+\bar{A} \cdot B$ Put $A=1$ and $B=1$, then $D=1 \cdot \overline{1}+\overline{1} \cdot 1=1 \cdot 0+0 \cdot 1=0+0=0$ So, this option is wrong. In option (b)
where, $D=\overline{[(\bar{A}+B)+(A+\bar{B}}]=[(\overline{\bar{A}+A)+(B+\bar{B})}]$ Put $A=1$ and $B=1$, then $D=[(\overline{\overline{1}}+1)+(1+\overline{1})]$ $=[\overline{(0+1)+(1+\overline{0})}]=\overline{[1+1]}=\overline{[1]}=0$ So, this option is wrong. In option (c)
where, $D$ $\begin{aligned} & =(A+B) \cdot(\bar{A}+\bar{B}) \\ & =A \cdot \bar{A}+A \cdot \bar{B}+B \cdot \bar{A}+B \cdot \bar{B} \\ & =0+A \cdot \bar{B}+B \cdot \bar{A}+0=A \cdot \bar{B}+\bar{A} \cdot B \end{aligned}$ Put $A=1$ and $B=1$, then $D=1 \cdot \bar{l}+\bar{l} \cdot 1=1 \cdot 0+0 \cdot 1=0+0=0$ So, this option is wrong. In option (d)
where, $D=A \cdot B+\bar{A} \cdot \bar{B}$ Put $A=1$ and $B=1$, then $D=1 \cdot 1+\overline{1} \cdot \overline{1}=1+0 \cdot 0=1+0=1$ So, this option is right.