If $W$ is the center of the circle then $I$ is the incenter of $\triangle R W Z$. Moreover, PRIZ is a rhombus. It follows that $P I$ is twice the inradius of a 1-1- $\sqrt{2}$ triangle, hence the answer of $2-\sqrt{2}$. So $L I=\sqrt{2}$. Alternatively, one can show (note, really) that the triangle $O I L$ is isosceles.