Let O and $\(\mathrm{O}^{\circ}\)$ be the centres of circles P and C respectively.

The length of $\(\mathrm{AB}=24\)$ and as $\(O O^{\prime}\)$ is perpendicular to AB , we get $\(\mathrm{AN}=\mathrm{NB}=12\)$.

If we join AO and AO ', we get two right triangles ANO and ANO ' respectively.






In triangle ANO , using Pythagoras theorem, we get $\(\mathrm{ON}=\sqrt{\mathrm{AO}^2-\mathrm{AN}^2}=\sqrt{15^2-12^2}=\sqrt{81}=9\)$ (Pythagoras theorem - Hypotenuse $\({ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ )

Similarly in triangle $\(\mathrm{ANO}^{\prime}\)$, we get $\(\mathrm{O}^{\prime} \mathrm{N}=\sqrt{\mathrm{AO}^{\prime 2}-\mathrm{AN}^2}=\sqrt{13^2-12^2}=\sqrt{25}=5\)$

So, the distance between the centres of two circles is $\(\mathrm{ON}+\mathrm{O}^{\prime} \mathrm{N}=9+5=14\)$.
Hence the answer is (D).