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There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that $$ \int_{B_R(0)}\|u\|_pd\mathbb{P}(u) > \frac{1}{2}\int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u). $$ Then if $x > B_{3R}(0)$, we see that $$ \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) > \int_{\mathbb{R}^n}\|u\|_pd\mathbb{P}(u) $$ and hence is $$ \inf_{x\in\mathbb{R}^n}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u) = \inf_{x\in B_{3R}(0)}\int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u). $$ Now, just note that $x\mapsto \int_{\mathbb{R}^n}\|u-x\|_pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B_{3R}(0)$. Moreover, we can say that the minimizer is unique since the function is convex (that follows from the convexity of the $p$-norm).