@nfdc23 answered this question in comments. Here is their comment as an answer, made CW to avoid reputation:

Sure, over any field $k$, via the "dynamic approach" to describing parabolic $k$-subgroups. Pick a 1-parameter $k$-subgroup $\lambda:\mathbf{G}_m\to P$ such that $P=P_G(\lambda)$, so $P=L \ltimes U$ for the smooth connected unipotent $U:=U_G(\lambda)$ and connected reductive $L:=Z_G(\lambda)$, so $U=\mathscr{R}_u(P)$ is the unipotent radical and $L$ is a Levi $k$-subgroup. The crucial point is that for $U^{-}:=U_G(-\lambda)$ the multiplication map $U^{-}\times L\times U=U^{-}\times P\to G$ is an open immersion. See section 2.1 of the book Pseudo-reductive Groups [by Conrad, Gabber, and Prasad] for a much wider context.