I labeled some points on your diagram.
Unfortunately I don't have a nice solution to your problem. Here is one solution.
Let y = |PA| and x = |BQ| then using similar triangles
y/4 = 4/x or x = 16/y
Also triangle POQ is a right triangle and hence Pythagoras Theorem gives
(y + 4)2 + (x + 4)2 = 202
Substitute x = 16/y and simplify. I got
y4 + 8 y3 - 368 y2 + 128y + 256 = 0
I then used Mathematica to solve the quartic equation and got four solutions which numerically are
y = 1.04, y = 15.35, y = -23.72 and y = -0.67
From the diagram the result you are looking for is y = 15.34 which symbolically is
y = 2(-1 + Sqrt + Sqrt[23 - 2 Sqrt]) where Sqrt is the square root.
Thus the length you want is
4 + 2(-1 + Sqrt + Sqrt[23 - 2 Sqrt])
which is approximately 19.35 meters.