Let $G$ be a simple algebraic group and let $P$ be a parabolic subgroup of $G$. It follows $X:=G/P$ is a smooth projective variety - the flag variety of $G$ corresponding to $P$.
Is it true that the following holds:
Pic($X$) has rank $1$ iff $P$ is a maximal parabolic subgroup.
Why? Where do i find a reference?
I'm not really sure what you're asking here--a maximal parabolic subgroup of $G$ is $G$ itself! That said, I'm fairly sure the below will answer your question.
Note that if one has a parabolic subgroup $P$ of $G$, a split connected group, then one has an exact sequence
$$0\to X^\ast(G)\to X^\ast(P)\to \mathrm{Pic}(G/P)\to \mathrm{Pic}(G)\to\mathrm{Pic}(P)\to 0$$
(e.g. see [Mil, §18.f]). From this, we see that if $G$ is semisimple then $X^\ast(G)=0$ and $\mathrm{Pic}(G)$ is finite (e.g. see [Mil, Corollary 18.23]) and thus this above says that we have an injection $X^\ast(P)\to \mathrm{Pic}(G/P)$ with finite quotient. Thus, the rank of $X^\ast(P)$ and $\mathrm{Pic}(G/P)$ are the same. But, since any map $P\to\mathbb{G}_m$ factors through $P/R_u(P)$ if $L$ is a Levi factor of $P$ we see that $X^\ast(P)=X^\ast(L)$. Moreover, we see that any map $L\to\mathbb{G}_m$ factors through $L^\mathrm{ab}$ and, in fact, the rank of $X^\ast(L)$ is the dimension of $L^\mathrm{ab}$ which is equal to the dimension of $Z(L)$ by standard theory (e.g. see [Mil, Example 19.25]).
Putting all this together we deduce the following: