Since $P$ is finite, it is clear that it will land inside the group over a finite subfield, say $\mathbb F_q$, $q$ a power of $p$. Then $P$ will, up to conjugation, lie in a Sylow $p$-subgroup.

So it seems that we want to know a description of the Sylow $p$-subgroups of $G(\mathbb F_q)$. In any matrix group, these are the maximal unipotent subgroups, since a subgroup is a $p$-group if and only if it is unipotent.

Using a faithful reprentation of the reductive group, we can see from the argument for $GL_n$ that the image of $P$ lies inside a unipotent subgroup. Thus it lies inside a maximal unipotent subgroup. So the correct analogue of the group of upper-triangular unipotent matrices is a maximal unipotent subgroup of the reductive group. (Since conversely, every unipotent element has order a power of $p$ and so generates a $p$-group, since this is true in $GL_n$.)

These are all conjguate since they are all the unique maximal unipotent subgroup of a Borel subgroup, all of which are conjugate. So the image of a $P$-group is contained, up to conjugation, in a specific maximal unipotent subgroup.