2. In the tetrahedron $P-ABC$, $PA \perp$ plane $ABC$, $AC \perp BC$. If $AC=2$, the dihedral angle $P-BC-A$ is $60^{\circ}$, and the volume of the tetrahedron $P-ABC$ is $\frac{4 \sqrt{6}}{3}$, then the sine value of the angle formed by line $PB$ and plane $PAC$ is
2. $\frac{\sqrt{3}}{3}$.
$$
\begin{array}{l}
\text { Given } A C \perp B C, P A \perp B C \\
\Rightarrow B C \perp \text { plane } P A C \\
\Rightarrow B C \perp P C .
\end{array}
$$

Since the dihedral angle $P-B C-A$ is $60^{\circ}$, we have,
$$
\begin{array}{l}
P A=2 \sqrt{3}, P C=4 . \\
\text { Given } \frac{4 \sqrt{6}}{3}=V_{\text {tetrahedron } P-A B C} \\
=\frac{1}{3} P A \cdot \frac{1}{2} A C \cdot B C \\
\Rightarrow B C=2 \sqrt{2} .
\end{array}
$$

Since $P A \perp$ plane $A B C$, the angle between line $P B$ and plane $P A C$ is $\angle B P C$. Therefore,
$$
P B=2 \sqrt{6}, \sin \angle B P C=\frac{\sqrt{3}}{3} \text {. }
$$