Let $$f(x)$$ be a quadratic polynomial with leading coefficient $$1$$ such that $$f(0) = p, p \neq 0$$, and $$f(1) = \dfrac{1}{3}$$. If the equations $$f(x) = 0$$ and $$fofofof(x) = 0$$ have a common real root, then $$f(-3)$$ is equal to ______.

Let $$f(x) = x^2 + bx + p$$ be a quadratic with leading coefficient 1, $$f(0) = p \neq 0$$, and $$f(1) = \frac{1}{3}$$.

From $$f(1) = 1 + b + p = \frac{1}{3}$$, we get $$b + p = -\frac{2}{3}$$.

If $$\alpha$$ is a common root of $$f(x) = 0$$ and $$f\circ f\circ f\circ f(x) = 0$$, then since $$f(\alpha) = 0$$:

$$f(f(\alpha)) = f(0) = p$$ $$f(f(f(\alpha))) = f(p) = p^2 + bp + p$$ $$f(f(f(f(\alpha)))) = f(p^2 + bp + p) = 0$$

So $$p^2 + bp + p$$ must be a root of $$f(x) = 0$$.

Note that $$p^2 + bp + p = p(p + b + 1) = p \cdot \frac{1}{3} = \frac{p}{3}$$.

Since $$\frac{p}{3}$$ is a root of $$f$$, we have

$$f\bigl(\frac{p}{3}\bigr) = \frac{p^2}{9} + \frac{bp}{3} + p = 0$$

Dividing by $$p\neq 0$$ gives $$\frac{p}{9} + \frac{b}{3} + 1 = 0$$, hence $$p + 3b + 9 = 0$$.

Solving the system $$b + p = -\frac{2}{3}$$ and $$p + 3b = -9$$: subtracting gives $$2b = -9 + \frac{2}{3} = -\frac{25}{3}$$, so $$b = -\frac{25}{6}$$. Then $$p = -\frac{2}{3} + \frac{25}{6} = \frac{21}{6} = \frac{7}{2}$$.

Finally, compute $$f(-3)$$:

$$f(-3) = 9 + \left(-\frac{25}{6}\right)(-3) + \frac{7}{2} = 9 + \frac{25}{2} + \frac{7}{2} = 9 + 16 = 25$$

The answer is $$\boxed{25}$$.