Let $L/K$ be a Galois extension with Galois group $G:=\mathrm{Gal}(L/K)$; suppose $K$ contains the $p$-th roots of unity, $p$ a prime number.
Now, if $\alpha \in L^\times$, $\overline{\alpha} \in H^0(G,L^\times/(L^\times)^p)$, then $L(\alpha^{1/p})/K$ is Galois, where $\overline{\alpha}$ denotes the class of $\alpha$ in $L^\times/(L^\times)^p$
What I know is that $L(\alpha^{1/p})/L$ is Galois with Galois group isomorphic to $\mathbb{Z}/p\mathbb{Z}$, but this doesn't guarantee me that $L(\alpha^{1/p})/K$ is Galois.
I keep all your notations, noting only that the $p$-th root $\alpha ^{1/p}$ is defined up to a $p$-th root of unity $\zeta$, but the cyclic extension $L(\alpha ^{1/p})$ does not depend on $\zeta$ which belongs to $L$. A classical criterion for $L/K$ to be Galois is that $L(\alpha ^{1/p})$ is stabilized by any extension $S$ of any $s\in G$ to a normal closure of $L(\alpha ^{1/p})$. By hypothesis $s(\alpha)=\alpha. \beta^p$, with $\beta \in L^*$ (this is just the explicitation of the invariance of $\bar\alpha$ under the action of $G$). So necessarily $S(\alpha ^{1/p})$ has the form $S(\alpha ^{1/p})=\alpha^{1/p}. \beta. \zeta $ , hence belongs to $L(\alpha ^{1/p})$, and we are done.