4. As shown in Figure 1, in $\triangle B C D$, $\angle B C D=90^{\circ}, B C=C D$ $=1, A B \perp$ plane $B C D, \angle A D B$ $=60^{\circ}$. Points $E$ and $F$ are on $A C$ and $A D$ respectively, such that plane $B E F \perp$ plane $A C D$, and $E F / / C D$. Then the sine value of the dihedral angle formed by plane $B E F$ and plane $B C D$ is ( ).
(A) $\frac{\sqrt{6}}{6}$
(B) $\frac{\sqrt{7}}{7}$
(C) $\frac{\sqrt{2}}{4}$
(D) $\frac{1}{3}$
4.B.

From the given, we know that the edge of the dihedral angle passes through point $B$ and is parallel to $CD$. Since $CD \perp$ plane $ABC$, $\angle CBE$ is the plane angle of the dihedral angle. Also, since $AE \perp EF$, it follows that $AE \perp$ plane $BEF$. Therefore, $AE \perp BE$. Thus, $\angle CBE = \angle BAC$. Furthermore, given $\angle BCD = 90^{\circ}$, $BC = CD$, and $\angle ADB = 60^{\circ}$, we can deduce that $AB = \sqrt{3} BD = \sqrt{6} BC$. Hence, $AC = \sqrt{7} BC$.