4. As shown in Figure 3, given a circle with radius 1 and center at $M(0,1)$, point $B(0,2)$, and point $A$ is on the negative half of the $x$-axis. $D$ is the midpoint of $O A$, and $A B$ intersects $\odot M$ at point $C$. If quadrilateral $B C D M$ is a parallelogram, then $\sin \angle A B D=$ $\qquad$
4. $\frac{\sqrt{10}}{10}$.

As shown in Figure 8, connect $O C$.
Then $\angle O C A=90^{\circ}$.
Therefore, quadrilateral $B C D M$ is a parallelogram
$$
\begin{array}{l}
\Rightarrow C D / / M B \Rightarrow C D \perp O A \\
\Rightarrow O C=O A \Rightarrow \angle C A O=45^{\circ} \\
\Rightarrow O A=O B \Rightarrow A D=1 .
\end{array}
$$

Draw $D E \perp A C$ at point $E$. Then $D E=\frac{\sqrt{2}}{2}$.
And $B D=\sqrt{5}$, so
$$
\sin \angle A B D=\frac{D E}{B D}=\frac{\sqrt{10}}{10} .
$$