1. In a regular quadrilateral pyramid $P-A B C D$, $\angle A P C=$ $60^{\circ}$. Then the cosine value of the dihedral angle $A-P B-C$ is $(\quad)$.
(A) $\frac{1}{7}$
(B) $-\frac{1}{7}$
(C) $\frac{1}{2}$
(D) $-\frac{1}{2}$
-、1.B.
As shown in Figure 4, within the side $PAB$, construct $AM \perp PB$, with the foot of the perpendicular being $M$. Connect $CM$ and $AC$, then $\angle AMC$ is the plane angle of the dihedral angle $A-PB-C$. Suppose $AB=2$, then $PA=AC=2\sqrt{2}$, and the slant height is $\sqrt{7}$.

Thus, $2 \times \sqrt{7} = AM \cdot 2\sqrt{2}$. From this, we get $CM = AM = \sqrt{\frac{7}{2}}$. Therefore, $\cos \angle AMC = \frac{AM^2 + CM^2 - AC^2}{2AM \cdot CM} = -\frac{1}{7}$.