jmc

Since $$\begin{array}{rl}p(p(1))& =p(1)=1,\\ p(p(2))& =p(3)=2,\\ p(p(3))& =p(2)=3,\end{array}$$three of the four solutions to $p(p(x))$ are $x=1,$ 2, and 3.

Also, the quadratic equation $p(x)=x$ has $x=1$ as a root. Let $r$ be the other root. Then $$p(p(r))=p(r)=r,$$so $r$ must be the fourth root we seek.

Since $p(x)-x=0$ for $x=1$ and $x=r,$ $$p(x)-x=c(x-1)(x-r)$$for some constant $c.$ Setting $x=2$ and $x=3,$ we get $$\begin{array}{rl}1& =c(2-r),\\ -1& =2c(3-r).\end{array}$$Dividing these equations, we get $$-1=\frac{2(3-r)}{2-r}.$$Solving for $r,$ we find $r=\boxed{\frac{8}{3}}.$