how can I solve the following equation for $p$? I am trying to solve the following equation for $p$ for a computational purpose.
$$\frac{\sqrt{1-(2ap+b)^2}(2ap+b) + \sin^{-1}(2ap+b)}{4a} = c$$
 A: You are, in effect, looking to solve the equation
$$\sin2x=8ac-2x$$
for $x$, with $-\pi/2\le x\le\pi/2$.  That is, we can simplify the original equation by getting rid of the arcsine if we let $2ap+b=\sin x$ with $-\pi/2\le x\le\pi/2$. That (and moving the $4a$ to the right hand side) gives us
$$\sqrt{1-\sin^2x}\sin x+x=4ac$$
we next notice that $\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=|\cos x|=\cos x$, the final equality relying on the restriction $-\pi/2\le x\le\pi/2$.  This gives $\cos x\sin x+x=4ac$, or $\sin x\cos x=4ac-x$. Multiplying both sides by $2$ and using $2\sin x\cos x=\sin2x$ gives the asserted equation.
The reason I write it in the given form is that the sine curve $y=\sin2x$ and the straight line $y=8ac-2x$ are relatively easy to sketch, and so it's relatively easy to see whether (and roughly where) they intersect.  In particular, if $0\le8ac\le\pi$, there will be at least one solution, and a modicum of calculus shows there will be exactly one solution in that case, and no solutions if $\pi\lt8ac$.  That's because the straight line passes through the $x$ axis at $4ac$ and has slope $-2$, which is matched by the derivative of the sine curve only at the endpoint, $x=\pi/2$.
Except for very special values of $8ac$ (such as $0$), there is no nice solution for the equation $\sin2x=8ac-2x$, so the best one can do, in general, is use numerical methods to get approximations.  One you do get a value for $x$, of course, you get what you want from 
$$p={\sin x-b\over2a}$$
