6. As shown in Figure 4, on side $BC$ of $\triangle ABC$, there is a point $D$, and $\angle ADB$ is an acute angle. $P$ and $Q$ are the circumcenters of $\triangle ABD$ and $\triangle ACD$, respectively, and the area of quadrilateral $APDQ$ is $\frac{3}{4}$ of the area of $\triangle ABC$. Then $\sin \angle ADB=$ $\qquad$
6. $\frac{\sqrt{6}}{3}$.

As shown in Figure 6, connect $P Q$ and $C Q$.
It is easy to prove that $\triangle A Q P \cong \triangle D Q P$.
Then $\frac{S_{\triangle P P}}{S_{\triangle A B C}}=\frac{3}{8}$.
Also, $\triangle A P Q \backsim \triangle A B C$
$$
\Rightarrow \frac{S_{\triangle A P Q}}{S_{\triangle A B C}}=\left(\frac{A Q}{A C}\right)^{2} \Rightarrow \frac{A Q}{A C}=\frac{\sqrt{3}}{2 \sqrt{2}} \text {. }
$$

Draw $Q H \perp A C$ at point $H$. Thus,
$$
\angle A D B=\frac{1}{2} \angle A Q C=\angle A Q H \text {. }
$$

Therefore, $\sin \angle A D B=\sin \angle A Q H=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$.