Let $G$ be a simply connected, semisimple algebraic group over $\mathbb C$ with maximal torus $T$ and Borel subgroup $B$ containing $T$. If $(V,\pi)$ is an irreducible representation of $G$, then $(V,d\pi)$ is an irreducible representation of the Lie algebra $\mathfrak g$ which has a unique highest weight $\lambda \in \mathfrak t^{\ast}$. I have read that if we identify the one-dimensional weight space $V_{\lambda} \subset V$ with a point in projective space $\mathbb P(V)$, then under the action $$ G \xrightarrow{\pi} \operatorname{GL}(V) \rightarrow \operatorname{Aut}(\mathbb P(V))$$ the stabilizer of $V_{\lambda}$ is a parabolic subgroup of $G$, and every parabolic subgroup of $G$ arises this way.

How does one go in the opposite direction? If $P$ is a (let's say maximal) parabolic standard subgroup of $G$, how does one find a dominant integral weight whose corresponding irreducible representation determines $P$ in the above sense? Can the highest weight occur as the highest root of $T$ in the unipotent radical of $P$?