4. As shown in Figure 4, in quadrilateral $ABCD$, $\angle ABC=135^{\circ}$, $\angle BCD=120^{\circ}$, $AB=\sqrt{6}$, $BC=5-\sqrt{3}$, $CD=6$. Then $AD=$
$$
\begin{array}{l}
CF=\frac{1}{2} CD=3, \\
DF=CD \sin 60^{\circ}=3 \sqrt{3}, \\
BE=AB \sin 45^{\circ}=\sqrt{3}, \\
AE=BE=\sqrt{3}, \\
DG=DF-FG=DF-AE=2 \sqrt{3}, \\
AG=EF=EB+BC+CF=\sqrt{3}+5-\sqrt{3}+3=8. \\
\text{Therefore, } AD=\sqrt{AG^{2}+DG^{2}}=\sqrt{8^{2}+(2 \sqrt{3})^{2}}=2 \sqrt{19}. \\
\end{array}
$$