Let $L$ be a normal extension over $k$, and let $a\in\overline{k}\cap L^c$ such that $L(a)$ is Galois over $L$. Then show there exists a finite extension $L'$ over $k$ such that $L'$ is contained in $L$, $L'(a)|L'$ is Galois and $Gal(L'(a)|L)$ is isomorphic to $Gal(L(a)|L)$. In particular if $L$ is Galois over $k$ then we can choose $L'$ Galois over $k$.
Since $L(a)|L$ is Galois, it is normal and separable and $min(a,L)$ splits in $L(a)$. I think I have to use the Galois correspondence here i.e. I need to get a proper subgroup of $Gal(L(a)|L)$ whose fixed field will be $L'$. But I cannot really think of it.
Maybe this can be useful. Let $a_1,\ldots,a_k$ the coefficients of the minimal polynomial of $\alpha$ over $L$. Set $L_1=k(a_1,\ldots,a_k)$. Let $M$ be the normal closure of $L_1(\alpha)$ and set $L^{\prime}:=M\cap L$. Notice that $M/L^{\prime}$ is Galois, this implies that $Gal(L(\alpha)/L)=Gal(ML/L)\cong Gal(M/L^{\prime})$. Since $L_1\subseteq L^{\prime}$, $[M:L^{\prime}]=[L^{\prime}(\alpha):L^{\prime}]$, so $L^{\prime}(\alpha)=M$.
$L^{\prime}$ should be the extension that you are looking for.