jmc

A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p{)}^2+p$. Because the $y$-intercept is $(0,-p)$ and $p\ne 0$, it follows that $a=-2/p$. Thus $$y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}{x}^2+4x-p,$$ so $\boxed{b=4}$.