Solution:

Let the roots of $p(x)$ be $r$ and $s$, so $$p(x)=(x-r)(x-s)=x^2-(r+s)x+rs.$$ 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 $$(r+s)^2-4(rs-r)=(r-s)^2+4r,$$ 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.