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
A. $\frac{1}{7}$
B. $-\frac{1}{7}$
C. $\frac{1}{2}$
D. $-\frac{1}{2}$
Given a regular quadrilateral pyramid $P-ABCD$ with a base edge length of 2, then its lateral edge length is $2\sqrt{2}$. As shown in the figure, draw $AE \perp PB$ at $E$, and connect $CE$. Then $\angle AEC$ is the plane angle of the dihedral angle $A-PB-C$. In $\triangle PAB$, $\cos B=\frac{1}{2\sqrt{2}} \Rightarrow \sin B=\frac{\sqrt{14}}{4}$
$$
\Rightarrow AE=\frac{\sqrt{14}}{2} \Rightarrow \cos \angle AEC=\frac{\frac{7}{2}+\frac{7}{2}-8}{2 \cdot \frac{7}{2}}=-\frac{1}{7} \text{. }
$$

Thus, the cosine value of the plane angle of the dihedral angle $A-PB-C$ is $-\frac{1}{7}$. Therefore, the answer is $B$.