help please

Denote M as the midpoint of AB, N as the midpoint of CD.

The answer would be sqrt(PM^2 + PN^2) by Pythagorean theorem.

Note that PM = (AB)/2 - PA = 3.

Let x = PD. Then by power of points, 

$x(13 - x) = 4(10)\\ x^2 - 13x + 40 = 0\\ x = 8\text{ or }x = 5$

Then PC = 8, PD = 5 or PC = 5, PD = 8.

In any case, PN = 13/2 - 5 = 3/2.

Distance from P to the center of the circle = $\sqrt{3^2 + \left(\dfrac32\right)^2} = \dfrac{3\sqrt5}2$