Let $$f(x) = (x - \alpha )(x - \beta )$$

It is given that $$f(0) = p \Rightarrow \alpha \beta = p$$

and $$f(1) = {1 \over 3} \Rightarrow (1 - \alpha )(1 - \beta ) = {1 \over 3}$$

Now, let us assume that, $$\alpha$$ is the common root of $$f(x) = 0$$ and $$fofofof(x) = 0$$

$$fofofof(x) = 0$$

$$ \Rightarrow fofof(0) = 0$$

$$ \Rightarrow fof(p) = 0$$

So, $$f(p)$$ is either $$\alpha$$ or $$\beta$$.

$$(p - \alpha )(p - \beta ) = \alpha $$

$$(\alpha \beta - \alpha )(\alpha \beta - \beta ) = \alpha \Rightarrow (\beta - 1)(\alpha - 1)\beta = 1$$ ($$\because$$ $$\alpha \ne 0$$)

So, $$\beta = 3$$

$$(1 - \alpha )(1 - 3) = {1 \over 3}$$

$$\alpha = {7 \over 6}$$

$$f(x) = \left( {x - {7 \over 6}} \right)(x - 3)$$

$$f( - 3) = \left( { - 3 - {7 \over 6}} \right)(3 - 3) = 25$$