Let $G$ be a compact connected and simply connected Lie group and $G^\mathbb{C}$ be the complexification of Lie group (with is diffeomorphic with $G^\mathbb{C}\cong T^\*G$) then I am looking for finding the 2nd homotopy group $\pi\_2(G^\mathbb{C}/P)$ where $P$ here is Parabolic subgroup. Is there any method or referrence 
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Finding the 2nd homotopy group $\pi_2(G^\mathbb{C}/P)$
As @abx has indicated, $\pi\_2(G\_{\mathbb{C}}/P)$ is isomorphic to $H\_2(G\_{\mathbb{C}}/P)$. To describe the latter group, we use the Bruhat decomposition of $G\_{\mathbb{C}}/P$. Choose a maximal torus $T$ and Borel $B$ satisfying $T\subseteq B\subseteq P$. We then have the Bruhat decomposition $$G\_{\mathbb{C}}/P=\coprod\_{[w]\in W/W\_P}BwP/P,$$ where $W\_P$ is the subgroup of $W$ generated by the reflections for the simple roots whose negative root spaces belong to the Lie algebra of $P$. The dimension of the Bruhat cell indexed by a coset in $W/W\_P$ is twice the length of the coset representative of minimal length. So, all cells are even-dimensional, and $H\_2(G\_{\mathbb{C}}/P)$ is the free abelian group of rank equal to the number of cosets in $W/W\_P$ with minimal representatives of length $1$. So, this rank is equal to the number of simple roots whose negative root spaces are not in the Lie algebra of $P$. (If $P=B$, then this rank is precisely the number of simple roots.) I hope this is the sort of answer you are seeking. 


