Let $\ell$ be the second tangent through $P$ to the circle aside from $BC$. Note that
$$\begin{aligned} \angle ABP &= 2\angle IBP = 2(90^\circ - \angle BPI) = 180^\circ - 2\angle BPI \\ &= 180^\circ - \angle (BP, \ell) = \angle (CP, \ell), \end{aligned}$$
where we use the fact that $BI$ is the angular bisector of $\angle ABP$, and that $PI$ is the angular bisector of the angle between $PB$ and $\ell$. It follows that $AB \parallel \ell$. The distance between these two parallel lines is $2r$ with $r$ the radius of the incircle.

Denote the distance from $C$ to $AB$ by $h$. Then we see that the area of $\triangle ABC$ equals $rs$ on the one hand, and $\frac{1}{2}ch$ on the other hand. It now follows from $rs = \frac{1}{2}ch$ that
$$\frac{|BC|}{|BP|} = \frac{d(AB, C)}{d(AB, \ell)} = \frac{h}{2r} = \frac{s}{c}.$$
We conclude that
$$\frac{|PC|}{|BP|} = \frac{|BC| - |BP|}{|BP|} = \frac{|BC|}{|BP|} - 1 = \frac{s}{c} - 1 = \frac{s-c}{c}. \quad \square$$