Isosceles triangle Let $ \triangle ABC $ be an $C$-isosceles and $ P\in (AB) $ be a point so that $ m\left(\widehat{PCB}\right)=\phi $. Express $AP$ in terms of $C$, $c$ and $\tan\phi$.
Edited problem statement(same as above but in different words):
Let $ \triangle ABC $ be a isosceles triangle with right angle at $C$. Denote $\left | AB \right |=c$. Point $P$ lies on $AB(P\neq A,B)$ and angle $\angle PCB=\phi$. Express $\left | AP \right |$ in terms of $c$ and $\tan\phi$. 

Edited for revised question
Dropping the perpendicular from $C$ onto $AB$ will help.  Call the point $E$.  
Also drop the perpendicular from $P$ onto $BC$, and call the point $F$.  Then drop the perpendicular from $F$ onto $AB$, and call the point $G$.  
This gives a lot of similar and congruent triangles.

$$\tan \phi = \dfrac{|PF|}{|CF|} = \dfrac{|FB| }{ |CF|} = \dfrac{ |GB| }{|EG| } = \dfrac{ |PB| }{|AP| }= \dfrac{ c-|AP| }{|AP| }$$ so $$|AP| =  \dfrac{c}{ 1+\tan \phi}.$$
