5. $-\sqrt{5}$.

Assume the edge length of the cube is $1, B P=x$, and let $A_{1} B$ intersect $B_{1} P$ at point $Q$.
It is easy to know that $Q B=\frac{\sqrt{2} x}{x+1}$.
Thus, the distance from point $B$ to the plane $B_{1} P C$ is
$h=Q B \sin 60^{\circ}=\frac{\sqrt{6} x}{2(x+1)}$.
Note that, in $\triangle B_{1} P C$,
$B_{1} C=\sqrt{2}, B_{1} P=P C=\sqrt{x^{2}+1}$.
Therefore, $S_{\triangle B_{1} P C}=\frac{\sqrt{2 x^{2}+1}}{2}$.
Applying the equal volume method to the tetrahedron $B_{1}-P B C$ we get
$V_{\text {tetrahedron } B_{1}-P B C}=V_{\text {tetrahedron } B-P B_{1}} C$.
Substituting in, we get
$\frac{x}{6}=\frac{x \sqrt{6\left(2 x^{2}+1\right)}}{12(x+1)} \Rightarrow x=\frac{1}{2}$.
Thus, the distance from point $B$ to $P B_{1}$ is $d=\frac{1}{\sqrt{5}}$.
Since the dihedral angle $A_{1}-B_{1} P-C$ is obtuse, the sine value is $\frac{h}{d}=\frac{\sqrt{5}}{\sqrt{6}}$, and the tangent value is $-\sqrt{5}$.
5. Given that $P$ is a point on the edge $AB$ of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, satisfying that the angle between line $A_{1} B$ and plane $B_{1} C P$ is $60^{\circ}$. Then the tangent value of the dihedral angle $A_{1}-B_{1} P-C$ is . $\qquad$