Consider a small length of the curve, $\mathrm{ds}$, then by Pythagoras ${\mathrm{ds}}^2={\mathrm{dx}}^2+{\mathrm{dy}}^2$

Rearrange this to give

${\displaystyle \frac{\mathrm{ds}}{\mathrm{dx}}=\sqrt{1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}}$

${\displaystyle \Rightarrow s={\int }_a^b\sqrt{1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}\mathrm{dx}}$

$s$ will be the length along the curve, since we are summing each small length element w.r.t. $x$.

We require that the we can find a closed form of the integral since we need a closed form function in terms of $p$ so that we can solve this to find $p$.

If we call the integral $f(x)$, then:

$f(p)=s+f(a)$

We already know $s$ and $f(a)$ by your starting conditions hence we only require that we can analytically find $f^{-1}(x)$, that is, the inverse function, such that $p=f^{-1}(s+f(a))$

If $f$ is something simple like $x^2$, then $f^{-1}(y)=y^{1/2}$ and we can find $p$ relatively easily.

In your example, we have $f(x)={\int }_a^p(1-{\cos }^{2}x{)}^{1/2}\mathrm{dx}={\int }_a^p\sin x\mathrm{dx}=-\cos p+\cos a$

So, $p={\cos }^{-1}[\cos a-s]$

If $f(x)$ is something like $\cos x+x$ then you'll find it very hard to solve this analytically - you can use the Newton-Raphson method or another iteration method to find the intersection of the line $y=s-\cos a-a-x$ and $y=\cos x$, the answer will be $p$... If you'd like me to explain how to do such iterative methods to find $p$ then just say and I'll write up the general idea.

I apologise if I covered anything too fast, I'm in a bit of a rush at the moment.

If there's anything you want me to reexplain then don't hesitate to reply.