If $r,s$ are roots of $p,$ solve $p(x)=r$ and $p(x)=s$

Exactly one of those quadratics must have discriminant $0$

The polynomial must have two distinct real roots: a repeated real root produces at most two real solutions of $p(p(x))=0,$ while nonreal roots produce none. Let its roots be $r$ and $s,$ so $$\begin{aligned} p(x) &=(x-r)(x-s) \\ &=x^2-(r+s)x+rs. \end{aligned}$$ The equation $p(p(x))=0$ is equivalent to $p(x)=r$ or $p(x)=s.$

For exactly three real solutions, one of these two quadratic equations must have a double root and the other must have two distinct real roots. Suppose $p(x)=r$ has the double root. Its discriminant is $$\begin{aligned} &(r+s)^2-4(rs-r) \\ &=(r-s)^2+4r, \end{aligned}$$ so $(r-s)^2=-4r,$ forcing $r\le0.$

The other equation, $p(x)=s,$ has discriminant $(r-s)^2+4s$ $=-4r+4s$ $=4(s-r),$ which must be positive. Hence $s\gt r,$ so $r-s=-2\sqrt{-r}$ and $s=r+2\sqrt{-r}.$

The sum of the roots is $r+s=2r+2\sqrt{-r}.$ Let $u=\sqrt{-r},$ so this is $-2u^2+2u,$ maximized at $u=\frac{1}{2}.$ Thus $r=-\frac{1}{4}$ and $s=\frac{3}{4}.$

Therefore $p(x)=x^2-\frac{1}{2}x-\frac{3}{16},$ and $$p(1)=1-\frac{1}{2}-\frac{3}{16}=\frac{5}{16}.$$

Thus, A is the correct answer.