Given, $f(x)=x^3+p x^2+q x$ is defined on $[0,2]$
$$
\begin{array}{ll}
f(0)=f(2) & \\
& 0=2^3+p(2)^2+q(2) \\
\Rightarrow & 4 p+2 q+8=0
\end{array}
$$


$$
\begin{aligned}
\text { Now } f^{\prime}(x) & =\frac{d}{d x}\left(x^3+p x^2+q x\right) \\
f^{\prime}(x) & =3 x^2+2 p x+q
\end{aligned}
$$

Given, $f^{\prime}\left(1+\frac{1}{\sqrt{3}}\right)=0$
$$
\begin{aligned}
& 3\left(1+\frac{1}{\sqrt{3}}\right)^2+2 p\left(1+\frac{1}{\sqrt{3}}\right)+q=0 \\
& \therefore 2 p\left(1+\frac{1}{\sqrt{3}}\right)+q+3\left(1+\frac{1}{3}+\frac{2}{\sqrt{3}}\right)=0
\end{aligned}
$$

Eq. (ii) Subtracting by Eq. (i), we get
$$
\begin{aligned}
& \Rightarrow \quad \frac{2 p}{\sqrt{3}}+\frac{6}{\sqrt{3}}=0 \\
& \Rightarrow \quad 2 P+6=0 \Rightarrow P=-3
\end{aligned}
$$

Substituting $P=-3$ in Eq. (i)
$$
\begin{aligned}
& \quad 2(-3)+q+4=0 \\
& \Rightarrow \quad q=2 \\
& \therefore p^2+q^2=(-3)^2+(2)^2=9+4 \\
& p^2+q^2=13
\end{aligned}
$$

Hence, option (a) is correct.