Since, $$p(x)$$ has a local maximum at $$x=1$$ and $$a$$ local minimum at $$x=3$$.

$$\begin{aligned} \text { Let } & p^{\prime}(x)=k(x-1)(x-3) \\ \Rightarrow & p^{\prime}(x)=k\left(x^2-4 x+3\right) \\ \Rightarrow & p(x)=k\left(\frac{x^3}{3}-\frac{4 x^2}{2}+3 x\right)+c \\ \Rightarrow & p(x)=k\left(\frac{x^3}{3}-2 x^2+3 x\right)+c \end{aligned}$$

Now, $$p(1)=6$$

$$\Rightarrow k\left(\frac{1^3}{3}-2(1)^2+3(1)\right)+c=6$$

$$\begin{array}{rrr} \Rightarrow & k\left(\frac{1}{3}-2+3\right)+c & =6 \\ \Rightarrow & \frac{4}{3} k+c & =6 \end{array}$$

Also, $$p(3)=2$$

$$\begin{aligned} \Rightarrow \quad & k\left(\frac{3^3}{3}-2(3)^2+3(3)\right)+c =2 \\ \Rightarrow \quad & k(0)+c =2 \\ \Rightarrow \quad & c =2 \end{aligned}$$

$$\begin{aligned} \text { Now, } \quad & \frac{4}{3} k+2 =6 \\ \Rightarrow \quad & \frac{4}{3} k =4 \end{aligned}$$

$$ \begin{array}{rlrl} \Rightarrow \quad & k =3 \\ \therefore \quad & p^{\prime}(0) =3\left(0^2-4(0)+3\right) \\ \qquad =9 \end{array}$$