7. Given a regular triangular prism $A B C-A_{1} B_{1} C_{1}$ with all 9 edges of equal length, $P$ is the midpoint of edge $C C_{1}$, and the dihedral angle $B-A_{1} P-B_{1}=\alpha$. Then $\sin \alpha=$ $\qquad$
7. $\frac{\sqrt{10}}{4}$.

Solution 1 As shown in Figure 1, establish a spatial rectangular coordinate system with the line containing $AB$ as the $x$-axis, the midpoint $O$ of line segment $AB$ as the origin, and the line containing $OC$ as the $y$-axis.

Let the edge length of the regular triangular prism be 2. Then
$$
\begin{array}{l}
B(1,0,0), B_{1}(1,0,2), \\
A_{1}(-1,0,2), P(0, \sqrt{3}, 1) . \\
\text { Hence } \overrightarrow{B A_{1}}=(-2,0,2), \overrightarrow{B P}=(-1, \sqrt{3}, 1), \\
\overrightarrow{B_{1} A_{1}}=(-2,0,0), \overrightarrow{B_{1} P}=(-1, \sqrt{3},-1) .
\end{array}
$$

Let the vectors perpendicular to the planes $B A_{1} P$ and $B_{1} A_{1} P$ be $m=\left(x_{1}, y_{1}, z_{1}\right), n=\left(x_{2}, y_{2}, z_{2}\right)$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
m \cdot \overrightarrow{B A_{1}}=-2 x_{1}+2 z_{1}=0, \\
m \cdot \overrightarrow{B P}=-x_{1}+\sqrt{3} y_{1}+z_{1}=0 ;
\end{array}\right. \\
\left\{\begin{array}{l}
n \cdot \overrightarrow{B_{1} A_{1}}=-2 x_{2}=0, \\
n \cdot \overrightarrow{B_{1} P}=-x_{2}+\sqrt{3} y_{2}-z_{2}=0 .
\end{array}\right.
\end{array}
$$

From this, we can set $m=(1,0,1), n=(0,1, \sqrt{3})$.
Therefore, $|\boldsymbol{m} \cdot \boldsymbol{n}|=|\boldsymbol{m}| \cdot|\boldsymbol{n}||\cos \alpha|$, that is
$$
\sqrt{3}=\sqrt{2} \times 2|\cos \alpha| \Rightarrow|\cos \alpha|=\frac{\sqrt{6}}{4} \text {. }
$$

Thus, $\sin \alpha=\frac{\sqrt{10}}{4}$.
Solution 2 As shown in Figure 2.
$$
\begin{array}{l}
P C=P C_{1}, \\
P A_{1}=P B .
\end{array}
$$

Let $A_{1} B$ and $A B_{1}$
intersect at point $O$. Then $P O \perp$
plane $A B B_{1} A_{1}$.
Also, $A_{1} B \perp A B_{1}$, so $O B_{1} \perp$ plane $P A_{1} B$.

Draw $O E \perp A_{1} P$ in plane $P A_{1} B$ through point $O$, with the foot of the perpendicular at $E$, and connect $B_{1} E$. Then $\angle B_{1} E O$ is the plane angle of the dihedral angle $B-A_{1} P-B_{1}$.
Let $A A_{1}=2$. It is easy to find that
$$
P B=P A_{1}=\sqrt{5}, A_{1} O=B_{1} O=\sqrt{2}, P O=\sqrt{3} \text {. }
$$

In the right triangle $\triangle P A_{1} O$,
$A_{1} O \cdot P O=A_{1} P \cdot O E \Rightarrow \sqrt{2} \times \sqrt{3}=\sqrt{5} O E$
$\Rightarrow O E=\frac{\sqrt{6}}{\sqrt{5}}$.
Also, $B_{1} O=\sqrt{2}$, so
$B_{1} E=\sqrt{B_{1} O^{2}+O E^{2}}=\frac{4 \sqrt{5}}{5}$.
Thus, $\sin \alpha=\sin \angle B_{1} E O=\frac{B_{1} O}{B_{1} E}=\frac{\sqrt{10}}{4}$.