Solution:

Because the right angle is at $C,$ segment $AB$ is a diameter of the circumcircle; let $O$ be its center, so the radius is $2.$ Let $\alpha = \angle ACP$ and extend $\overline{CP}$ to meet the circle again at $D.$ The central angle over arc $AD$ is $\angle AOD = 2\angle ACD = 2\alpha,$ while vertical angles give $\angle DPB = \angle APC = 2\alpha.$ So $\overline{OD}$ and $\overline{PD}$ make equal angles with line $AB,$ and triangle $ODP$ is isosceles with $DP = DO = 2.$

By the power of the point $P,$ $$AP \cdot PB = CP \cdot PD = 1 \cdot 2 = 2, \qquad AP + PB = 4,$$ so $AP$ and $PB$ are the roots of $t^2 - 4t + 2,$ namely $2 \pm \sqrt{2}.$ Since $\angle BAC \lt 45^\circ,$ we have $BC \lt AC$ with $AC^2 + BC^2 = 16,$ so $AC \gt 2\sqrt{2},$ and the triangle inequality gives $AP \ge AC - CP \gt 2\sqrt{2} - 1 \gt 2 - \sqrt{2}.$ Hence $AP = 2 + \sqrt{2}.$

Therefore $$\frac{AP}{BP} = \frac{2 + \sqrt{2}}{2 - \sqrt{2}} = \frac{(2 + \sqrt{2})^2}{2} = 3 + 2\sqrt{2},$$ and $p + q + r = 3 + 2 + 2 = 7.$