Let $f(x)=x^2+bx+p$ be the quadratic polynomial, since it is monic and $f(0)=p$.

We are also given $f(1)=\frac13$ so 1+b+p=\frac13 \implies b+p=-\frac23. \tag{1}

We are told that the equations $f(x)=0$ and $f\circ f\circ f\circ f(x)=0$ have a common real root.

1. Interpret the common root condition

Let $\alpha$ be a common real root. Then $f(\alpha)=0$ and also $f^{(4)}(\alpha)=0,$ where $f^{(4)}=f\circ f\circ f\circ f$.

Since $f(\alpha)=0$, $f^{(4)}(\alpha)=f(f(f(f(\alpha))))=f(f(f(0))).$ Thus the condition becomes $f(f(f(0)))=0.$ But $f(0)=p$, so this is f(f(p))=0. \tag{2}

Now, the roots of $f$ are exactly the numbers sent to $0$ by $f$. So (2) says that $f(p)$ is a root of $f$.

Also, note that $f(p)=p^2+bp+p=p(p+b+1).$ Using (1), $b+1=\frac13-p,$ so $p+b+1=\frac13.$ Hence f(p)=p\cdot \frac13=\frac p3. \tag{3}

Therefore $\frac p3$ is a root of $f$.

2. Use the root condition

Since $\frac p3$ is a root, we must have $f\left(\frac p3\right)=0.$ Now $f\left(\frac p3\right)=\left(\frac p3\right)^2+b\left(\frac p3\right)+p.$ Using (1), $b=-\frac23-p.$ So \begin{align*} f\left(\frac p3\right) &=\frac{p^2}{9}+\frac p3\left(-\frac23-p\right)+p \ &=\frac{p^2}{9}-\frac{2p}{9}-\frac{p^2}{3}+p. \end{align*} Multiply by $9$: $p^2-2p-3p^2+9p=0$ $-2p^2+7p=0$ $p(7-2p)=0.$ Given $p\ne 0$, we get $p=\frac72.$

Then from (1), $b=-\frac23-\frac72=-\frac{25}{6}.$

So $f(x)=x^2-\frac{25}{6}x+\frac72.$

3. Compute $f(-3)$

\begin{align*} f(-3) &=(-3)^2-\frac{25}{6}(-3)+\frac72 \ &=9+\frac{25}{2}+\frac72 \ &=9+14+2 \ &=25. \end{align*}

4. Final answer

$\boxed{25}$

This matches the stored correct answer.