A Geometry problem!

$\overline{QS}$ is the perpendicular bisector of $\overline{BD} \implies \overline{OQ}$ is a radii of the circle with center $O$ .

Using the Pythagorean theorem $\implies \overline{CB} = p^2 + 1$ and $m\angle{A} = 90^{\circ} \implies$

$m(\widehat{CDB}) = m(\widehat{CAB}) = 180^{\circ} \implies \overline{CB}$ is a diameter of the above circle $\implies$

$R = \dfrac{\overline{CB}}{2} = \dfrac{p^2 + 1}{2} \implies \overline{OS} = R - 1 \implies \overline{BS}^2 = R^2 - (R - 1)^2 = 2R - 1 =$

$p^2 \implies \overline{BS} = p \implies \overline{BD} = 2p = \overline{AP}$

$\implies \triangle{ABD}$ is an isosceles triangle $\implies \overline{BE}$ is the $\perp$ bisector of $\overline{AD} \implies$

$\overline{AD} = 2\overline{AE}$

$\triangle{ABC} \implies \sin(\theta) = \dfrac{p^2 - 1}{p^2 + 1}$ and $\cos(\theta) = \dfrac{2p}{p^2 + 1} \implies$

$\overline{EB} = 2p\cos(\theta) = \dfrac{4p^2}{p^2 + 1}$ and $\overline{AE} = 2p\sin(\theta) = \dfrac{2p(p^2 - 1)}{p^2 + 1} \implies$

$\overline{AD} = \dfrac{4p(p^2 - 1)}{p^2 + 1}$

$\implies \triangle{ADB} = \dfrac{1}{2}(\overline{AD} * \overline{EB}) = \dfrac{8p^3(p^2 - 1)}{(p^2 + 1)^2}$

and

$\triangle{ABC} = p(p^2 - 1) \implies \dfrac{\triangle{ADB}}{\triangle{ABC}} = \dfrac{8p^2}{(p^2 + 1)^2} = \dfrac{8}{5}$

$\implies 5p^2 = p^4 + 2p^2 + 1 \implies p^4 - 3p^2 + 1 = 0 \implies$

$p^2 = \dfrac{3 \pm \sqrt{5}}{2} = (\dfrac{1 \pm \sqrt{5}}{2})^2 \implies p = \pm(\dfrac{1 \pm \sqrt{5}}{2})$

and $p > 1 \implies p = \phi = \dfrac{1 + \sqrt{5}}{2} =\dfrac{a + \sqrt{b}}{c} \implies a + b + c = \boxed{8}$ .