Let $G$ be a Lie Group and $\mathfrak{g}$ be its lie algebra.
Let $\mathfrak{g}$ is semisimple or reductive lie algebra, then prove that $\mathfrak{g}^\*$ (dual of $\mathfrak{g}$)is invariant under $Ad(G)$? 



when $g^*$ is invariant under $Ad(G)$?
The question you want to ask, in order to understand remark 1, page 3 of Kirillov's book, is why, if $\mathfrak{g}$ is a reductive or semisimple Lie group, for every representation of $\mathfrak{g}$, every $\mathfrak{g}$-invariant subspace has a $\mathfrak{g}$-invariant complement. In particular, Kirillov is assuming that some reductive or semisimple Lie algebra $\mathfrak{g}$ is faithfully represented in vector space $W=\mathbb{R}^{n \times n}$ of matrices. Then $\mathfrak{g} \to W$ is a $\mathfrak{g}$-equivariant linear injection. The dual map $W^\* \to \mathfrak{g}^\*$ is therefore a surjection, and has some kernel $\mathfrak{g}^{\perp}$. If $\mathfrak{g}^{\perp}$ has an invariant complement in $W$, we can identify that complement with $\mathfrak{g}^\*$ by the projection $W \to \mathfrak{g}^\*$. There are various definitions of reductive Lie algebras, but one of them is that a reductive Lie algebra is one all of whose finite dimensional representations split into a sum of irreducible representations. It is a theorem that every semisimple Lie algebra has this property. Therefore if $\mathfrak{g}$ is semisimple, or (more generally) reductive, then there is such a complement, which identifies $\mathfrak{g}^\*$ with a subspace of $W^\*$. The Killing form identifies $W$ with $W^\*$, so thereby identifies $\mathfrak{g}^\*$ with a linear subspace of matrices.


