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Assuming $C(L,A)$ positive, your inequality is not true in general. In fact you can have $\int_{\partial\Omega} p^2 dS >\int_{\partial\Omega^*} (p^*)^2 dS^*$. Indeed, if I am not mistaken, $$ \int_{\partial\Omega} p^2 dS = \int_0^{2\pi} \frac{r^4}{\sqrt{r^2+{r'}^2}} d\theta $$ where $r=r(\theta)$ is the radial function and $\theta$ is the angular coordinate. Begin with $\Omega^*$ being a polygon approximating a long and narrow ellipse. Its radial function $r^*(\theta)$ is piecewise smooth. Approximate $r^*$from below by a piecewise constant function so that the difference is bounded by a small $\varepsilon$. Near points of discuntinuity, connect the constant pieces by linear functions with almost infinite derivative. The resulting function is $r(\theta)$. If the approximation is chosen so that $r=r^*$ at points corresponding to vertices of the original polygon $\Omega^*$, then $\Omega^*$ is indeed the convex hull of the set $\Omega$ defined by $r(\theta)$. However, $$ \int_{\partial\Omega} p^2 dS \approx \int_0^{2\pi} r^3 d\theta\ge \int_0^{2\pi} (r^*-\varepsilon)^3 d\theta > \int_0^{2\pi} \frac{(r^*)^4}{\sqrt{(r^*)^2+{(r^*)'}^2}} = \int_{\partial\Omega^*} (p^*)^2 dS^* $$ for $\varepsilon$ sufficiently small, since the term ${(r*)'}^2$ in the denominator is not negligible.