6. In the convex quadrilateral $A B C D$, $\angle B=\angle D=$ $90^{\circ}, \angle C=120^{\circ}, A B=3, B C=\sqrt{3}$. Then $A D=$ ( ).
(A) $\sqrt{3}$
(B) 3
(C) $2 \sqrt{3}$
(D) $3 \sqrt{3}$
6. B.

As shown in Figure 3, extend $BC$ and $AD$ to intersect at point $E$. Since $\angle BCD=120^{\circ}$, we have
$$
\begin{array}{l}
\angle ECD=60^{\circ} \Rightarrow \angle E=30^{\circ} \\
\Rightarrow EA=2AB=6 \\
\Rightarrow BE=\sqrt{6^{2}-3^{2}}=3\sqrt{3}
\end{array}
$$

Also, $BC=\sqrt{3} \Rightarrow CE=2\sqrt{3} \Rightarrow ED=\frac{\sqrt{3}}{2}CE=3$. Therefore, $AD=3$.