Let $q_1,\dots,q_s$ be the vertices of $P$; denote by $o$ the center of $S$.
Note that $q_i\mapsto p_i$ is a distance-nonexpanding map. By Kirszbraun theorem it can be extended to a distance-nonexpanding map $f$ defined on the whole space. It follows that $p_1,\dots,p_s$ lie on the distance $\le r$ from $f(o)$. So the radius of $\{p_1,\dotsc,p_s\}$ is at most $r$.
Since the circumcenter of $\{p_1,\dotsc,p_s\}$ lies in $\operatorname{conv}\{p_1,\dotsc,p_s\}$, the circumradius $=$ radius of $\{p_1,\dotsc,p_s\}$. Whence the statement follows.