Consider the Quadrilateral $BTCS$ .Obviously it is harmonic Hence the cross ratio $(A,P,B,C)$ is harmonic. $\Longrightarrow \frac{AB}{BP}=\frac{AC}{CP} \Longrightarrow AC= \frac{PC.AB}{BP} \Longrightarrow \frac{AC}{AP}= \frac{PC.AB}{AP.BP} \Longrightarrow 1+\frac{PC}{AP}=PC.(\frac{1}{BP}-\frac{1}{AP}) \Longrightarrow 2+2.\frac{PC}{AP}=\frac{BC}{BP} \Longrightarrow 2.\frac{AC}{AP}=\frac{BC}{BP}$ and we have the result

1) Ptolemy for degenerated cyclic $ABPC$: $PC\cdot AB+PB\cdot AC=BC\cdot AP$ 2) $AS$ being tangent, $\frac{AB}{AC}=\left( \frac{BS}{CS}\right)^2$ 3) $SP$ being symmedian, it yields $\left( \frac{BS}{CS}\right)^2=\frac{PB}{CP}$. From $2) \wedge 3)$ we get $PB\cdot AC=PC\cdot AB$; with that, from $1)$ we get the desired relation. Best regards, sunken rock

Another solution, which is quick and easy, and I think it should work: Make a projective transformation $p$ that sends the circle to the circle and $P$ to the center of the circle. Let's introduce some notation to make it more readable: Call the circle $\omega$ with center $O$, and call the point at infinity $Q_{\infty}$. We make the projective transformation: $p(\omega)=\omega$ and $p(P)=O$. Under this transformation the tangents become parallel, and therefore $p(A)=Q_{\infty}$ Let's see what we must prove: $\frac{AP}{PC}=2 \cdot \frac{AB}{BC} \Leftrightarrow\frac{\frac{AP}{PC}}{\frac{AB}{BC}}=2$ We notice that $\frac{\frac{AP}{PC}}{\frac{AB}{BC}}=(A,C;P,B)$. Since the cross-ratio is preserved under projective transformations It is enough to prove that $(Q_{\infty},C;O,B)=2=\frac{\frac{Q_{\infty}P}{OC}}{\frac{Q_{\infty}B}{BC}}=\frac{Q_{\infty}P}{Q_{\infty}B} \cdot \frac{BC}{OC}$ $\frac{Q_{\infty}P}{Q_{\infty}B}=1$, as both are infinite lengths of same "type". $\frac{BC}{OC}=2$, obviously, as O is the circumcenter. Thus, $2=\frac{Q_{\infty}P}{Q_{\infty}B} \cdot \frac{BC}{OC}=(Q_{\infty},C;O,B)=(A,C;P,B)=\frac{\frac{AP}{PC}}{\frac{AB}{BC}}$ We are thus done. $\blacksquare$