Answer: 57

Solution:

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{CD}.$ The segment from the center to a chord's midpoint is perpendicular to the chord, so $OM = \sqrt{25^2 - 15^2} = 20$ and $ON = \sqrt{25^2 - 7^2} = 24,$ with $MN = 12.$

Since $P$ lies on both chords, $\angle OMP = \angle ONP = 90^\circ,$ so $M$ and $N$ lie on the circle with diameter $OP.$ In triangle $OMN,$ the law of cosines gives $$\begin{aligned} \cos \angle MON &= \frac{20^2 + 24^2 - 12^2}{2 \cdot 20 \cdot 24} \\ &= \frac{832}{960} = \frac{13}{15}, \end{aligned}$$ so $\sin \angle MON = \frac{2\sqrt{14}}{15}.$ In the circle through $O,$ $M,$ $P,$ $N,$ the extended law of sines says the chord $MN$ equals the diameter $OP$ times $\sin \angle MON,$ so $$\begin{aligned} OP &= \frac{12}{\frac{2\sqrt{14}}{15}} = \frac{90}{\sqrt{14}}, \end{aligned}$$ $$\begin{aligned} OP^2 &= \frac{8100}{14} = \frac{4050}{7}. \end{aligned}$$

Then $m + n = 4050 + 7 = 4057,$ which leaves remainder $57$ upon division by $1000.$