About structure of parabolic subgroups of finite classical algebraic groups I am interested about a Fact (if it is right) of the structure of parabolic subgroups of finite classical algebraic groups:
Let $G$ be a classical algebraic group over the finite field of order $r^{e}$, and let $P \leq G$ denote a maximal parabolic subgroup of $G$.
We regard the unipotent radical of $P$ by $ R:=R_{U}(P) = O_{r}(P)$. Is it true, that $R$ is a minimal normal subgroup of $P$?
(For $P$ being a parabolic subgroup which is not a maximal in $G$, it is not true.  For example, let $G:=\operatorname{PSL}(3,r^{e})$ and take as $P$ a Borel subgroup.)
Is it correct for the maximal parabolic subgroups of $G$? Are there any Theorems or Propositions about that?
 A: The standard reference for this is Azad, Barry, Seitz.
On the structure of parabolic subgroups.
Comm. Algebra 18 (1990), no. 2, 551–562.
It gives explicit information on the structure of the unipotent radical for the algebraic group case together with all split Chevalley groups and twisted 'Steinberg' groups, and also gives details on what happens when the characteristic is 'special', e.g. p=3, G=G_2.
A: The unipotent radical $U$ of a maximal parabolic $P$ of a classical group is not always abelian - see Jim's answer to a specific example and the paper by Richardson, Rohrle and Steinberg for the general case. So in these cases $Z(U)$ is proper subgroup of $U$ that is normal in $P$, and you have a counterexample.
In the cases where $U$ is abelian, one needs to check whether the natural action of a Levi subgroup of $P$ on the unipotent radical $U$ is irreducible. There are lots of sources for this sort of thing, for instance Volume 3 of the series by Gorenstein, Lyons and Solomon.
Edit: As I mention in my comment above on Jim's answer, I don't know under what circumstances $Z(U)$ is minimal normal - again this would come down to studying the irreducibility of the action of the Levi on $Z(U)$.
(I should add, in case anyone thinks I was cheeky, that I initially phrased this answer speculatively... and adjusted it in light of the counter-example given in  Jim's answer.)
A: No, you can see the problem here already in type $B_2$.  If the simple roots are called $\alpha$ (long) and $\beta$ (short), a maximal (proper) subgroup $P$ has as Levi factor a copy of $\mathrm{GL}_2(k)$ corresponding to the root $\alpha$, while its unipotent radical involves three root groups for $\beta, \alpha+\beta, \alpha+2 \beta$.  But the last of these root groups is a minimal normal subgroup of $P$.  (Here it also doesn't matter what the field $k$ is.  And I don't see why you specify "classical" types in the question.)
A: It is a general fact that if $P$ is a parabolic $k$-subgroup of an arbitrary connected reductive group $G$ over an arbitrary field $k$ then $U := \mathscr{R}_u(P)$ has a canonical $P$-equivariant filtration $$U = U_0 \supset U_1 \supset \dots \supset U_m \supset U_{m+1} = \{1\}$$ by smooth connected $k$-subgroups such that each $U_i/U_{i+1}$ is a vector group that admits a unique $P$-equivariant linear structure. (This can be defined over $\mathbf{Z}$ as well when working with "Chevalley groups", so it is truly characteristic-free). In particular, the final stage of this canonical filtration is a vector group that is a normal $k$-subgroup of $P$.  This must be well-known in the classical case, but the only reference I know in the literature is SGA3, Exp. XXVI, 2.1.
If $G$ is split semisimple of adjoint type and $T$ is a split maximal $k$-torus in $P$ then the resulting linear representation of $T$ on each $U_i/U_{i+1}$ is a direct sum of 1-dimensional weight spaces given by certain roots in $\Phi(G,T)$.  I expect that when the root system is moreover irreducible then each $(U_i/U_{i+1})(k) = U_i(k)/U_{i+1}(k)$ is irreducible as a $k$-linear $P(k)$-representation, apart from perhaps some special cases with low rank and/or $k$ with size 2 or 3.  In particular, the final stage $U_m(k)$ would then be minimal as a normal subgroup of $P(k)$ (since there are no divisible roots in the character lattice due to $G$ being of adjoint type).  Maybe this is sufficient for your purposes?  (No motivation was given for the posted question, beyond perhaps curiosity.)
Since you use the notation "PSL" which is a perennial source of confusion (do you mean ${\rm{SL}}_n(k)/\mu_n(k)$ or ${\rm{PSL}}_n(k)$ where ${\rm{PSL}}_n$ is meant as the quotient ${\rm{SL}}_n/\mu_n$ in the sense of algebraic groups?), I should note that although such confusion creates headaches with the tori, it does not with the "unipotent radicals", so the above should be applicable to various possible meanings of the phrase "finite classical group", but (as always) be careful.
