Assume the edge length of $A B C-A_{1} B_{1} C_{1}$ is $2$, and $E$ is the midpoint of $A_{1} B$. Draw $E F \perp A_{1} P$, and connect $P E, B_{1} P, B_{1} E, B_{1} F$.

Then $P A_{1}=P B=\sqrt{5} \Rightarrow P E=\sqrt{3} \Rightarrow B_{1} E \perp P E$ $\Rightarrow B_{1} E \perp$ plane $A_{1} P B$, so $\angle E F B_{1}$ is the plane angle of the dihedral angle $B-A_{1} P-B_{1}$. In $\triangle P E A_{1}$, $E F=\frac{A_{1} E \cdot P E}{P A_{1}}=\frac{\sqrt{30}}{5} \Rightarrow B F=\sqrt{E F^{2}+B_{1} E^{2}}$ $=\frac{4 \sqrt{5}}{5} \Rightarrow \sin \alpha=\sin \angle E F B_{1}=\frac{B_{1} E}{B F}=\frac{\sqrt{10}}{4}$.
7. In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, all 9 edges are of equal length, $P$ is the midpoint of $C C_{1}$, and the dihedral angle $B-A_{1} P-B_{1}=\alpha$, then $\sin \alpha=$ $\qquad$ .