China Mathematical Competition

Let the line through segment $AB$ be $x$-axis with the origin $O$ being the midpoint of $AB$ and let the line through segment $OC$ be $y$-axis to establish a space rectangular coordinate system shown in Fig. 7.1. Assuming the length of an edge is 2, we have $B(1, 0, 0)$, $B_1(1, 0, 2)$, $A_1(-1, 0, 2)$, $P(0, \sqrt{3}, 1)$. Then

Fig. 7.1
$$\begin{aligned} \overrightarrow{BA_1} &= (-2, 0, 2), \quad \overrightarrow{BP} = (-1, \sqrt{3}, 1), \\ \overrightarrow{B_1A_1} &= (-2, 0, 0), \quad \overrightarrow{B_1P} = (-1, \sqrt{3}, -1). \end{aligned}$$
Let vectors $\vec{m} = (x_1, y_1, z_1)$ and $\vec{n} = (x_2, y_2, z_2)$ be perpendicular to $BA_1P$ and $B_1A_1P$, respectively. We have
$$\begin{cases} \vec{m} \cdot \overrightarrow{BA_1} = -2x_1 + 2z_1 = 0, \\ \vec{m} \cdot \overrightarrow{BP} = -x_1 + \sqrt{3}y_1 + z_1 = 0, \\ \vec{n} \cdot \overrightarrow{B_1A_1} = -2x_2 = 0, \\ \vec{n} \cdot \overrightarrow{B_1P} = -x_2 + \sqrt{3}y_2 - z_2 = 0. \end{cases}$$
We can then assume that $\vec{m} = (1, 0, 1)$, $\vec{n} = (0, 1, \sqrt{3})$.
From $|\vec{m} \cdot \vec{n}| = |\vec{m}| \cdot |\vec{n}| \cos \alpha$, we have
$$\sqrt{3} = \sqrt{2} \cdot 2 \cos \alpha \Rightarrow \cos \alpha = \frac{\sqrt{6}}{4}.$$

As seen in Fig. 7.2, we have $PC = PC_1$, $PA_1 = PB$. Suppose $A_1B$ and $AB_1$ intersect at $O$. Then we get $OA_1 = OB$, $OA = OB_1$, $A_1B \perp AB_1$.
Since $PA = PB_1$, then $PO \perp AB_1$. Therefore $AB_1 \perp$ plane $PA_1B$.
On plane $PA_1B$ through $O$ draw line $OE \perp A_1P$ with foot point $E$.

Fig. 7.2
Connecting $B_1E$, $\angle B_1EO$ is then the plane angle of the dihedral angle $B-A_1P-B_1$. Assuming $AA_1 = 2$, it is easy to find that $PB = PA_1 = \sqrt{5}$, $A_1O = B_1O = \sqrt{2}$, $PO = \sqrt{3}$.
In right triangle $\triangle PA_1O$, we have $A_1O \cdot PO = A_1P \cdot OE$,
or $\sqrt{2} \cdot \sqrt{3} = \sqrt{5} \cdot OE$. So $OE = \frac{\sqrt{6}}{\sqrt{5}}$.
As $B_1O = \sqrt{2}$, we have
$$B_1E = \sqrt{B_1O^2 + OE^2} = \sqrt{2 + \frac{6}{5}} = \frac{4\sqrt{5}}{5}.$$
Finally,
$$\sin \alpha = \sin \angle B_1EO = \frac{B_1O}{B_1E} = \frac{\sqrt{2}}{\frac{4\sqrt{5}}{5}} = \frac{\sqrt{10}}{4}. \quad \square$$