The answer is negative: the area of $P(S)$ is at most the area of the unit sphere, while the area of $S$ can be made arbitrarily high by taking a small sphere with a lot of thin ``fingers''.

The answer is negative: the area of $P(S)$ is at most the area of the unit sphere, while the area of $S$ can be made arbitrarily high.

An $S$ contained in the unit sphere and star-shaped at $0$ can be parametrized by the radius in polar coordinates: $S=\{\phi(u)u : \lVert u\rVert=1\}$ where $\phi$ is any smooth function from the unit sphere to $(0,1)$. Now, the area of $S$ is something like $$\int \phi^{n-1}\sqrt{\lVert \nabla \phi\rVert^2+1}$$ (a bit late here, so I might have gotten the formula wrong but in any case the integrand goes to infinity with $\nabla\phi$). Taking $\phi$ with value in say $[\frac13,\frac23]$ and with a lot of variation (e.g. making fingers or wrinkles) we can easily make the area of $S$ arbitrarily high.