Complex bash with the incircle of $\triangle ABC$ as the unit circle, and let it touch sides $BC,CA,AB$ at $D,E,F$ respectively. Let $j$ denote the coordinate of $I$, so that \begin{align*} |d|=|e|=|f|&=1 \\ a &= \frac{2ef}{e+f} \\ b &= \frac{2df}{d+f} \\ c &= \frac{2de}{d+e} \\ j &= 0 \\ o &= \frac{2def(d+e+f)}{(d+e)(d+f)(e+f)} \end{align*}Now $P$ is the reflection of $O$ over the perpendicular bisector of $\overline{AI}$. This perpendicular bisector has equation $$\frac{z-\frac{a+j}2}{a-j} \in i\mathbb{R}$$$$\frac{ez+fz-ef}{2ef} = -\frac{e\overline{z}+f\overline{z}-1}2$$$$ez+fz+e^2f\overline{z}+ef^2\overline{z} = 2ef$$$$z = \frac{2ef}{e+f} - ef\overline{z}$$Then $$p = \frac{2ef}{e+f} - ef\overline{o} = \frac{2ef}{e+f} - \frac{2ef(de+df+ef)}{(d+e)(d+f)(e+f)} = \frac{2d^2ef}{(d+e)(d+f)(e+f)}$$Similarly, $$q = \frac{2de^2f}{(d+e)(d+f)(e+f)}$$Then \begin{align*} c-p &= \frac{2de(de+ef+f^2)}{(d+e)(d+f)(e+f)} \\ c-q &= \frac{2de(de+df+f^2)}{(d+e)(d+f)(e+f)} \end{align*}Then since $$c-q = de(\overline{c}-\overline{p})$$we have $|c-p| = |c-q|$. $\blacksquare$