Let $p$ be a prime number. Let $K$ be a finite extension of $\mathbb{Q}_p$. Let $a\in K^\times$ which is not a root of unity. Then do we have the following?
There exists $N$ such that for $n\geq N$, $K(\alpha)/K$ is not Galois for every $\alpha\in \overline{\mathbb{Q}}_p$ satisfies $\alpha^{p^n}=a$.
EDIT. Rewritten answer after discussion with the OP.
Now that I understand correctly your question, it seems that the solution is merely Galois theoretic. For simplification, suppose that $K$ contains $\mu_p$ and $p\neq 2$. Denote $K_n = K(\mu_{p^n}), G_n=Gal(K_n/K)$. Fix $a\in K^*$, not a $p$-primary root of 1, and define $L_n=K_n(\sqrt [p^n]a), H_n=Gal(L_n /K_n)$ (which is cyclic of order dividing $n$).
Obviously $L_n /K$ is normal iff the extensions of the elements of $G_n$ to automorphisms of a normal closure of $L_n$ stabilize $H_n$, and this action (via inner automorphisms) will define a character $\chi_n : G_n \to H_n$. But, by Kummer theory, $H_n\cong Hom (<a>_n,\mu_{p^n})$, where $<a>_n$ is the subgroup of ${K_n}^*/{{K_n}^*}^{p^n}$ generated by the class of $a$. Recall that the (only) functorial action of $G_n$ on $Hom(.)$ is defined by $(g(f))(x)=g(f(g^{-1}(x))$. It will be convenient to introduce here the notion of "Tate twist". The action of $G_n$ on $\mu_{p^n}$ defines the so called cyclotomic character $\kappa_n : G_n \to \mu_{p^n}$. By Pontryagin duality, $<a>_n\cong Hom (H_n, \mathbf Z/p^n)$, but this is not an isomorphism of $G_n$-modules because $G_n$ acts trivially on $\mathbf Z/p^n$, and via $\kappa$ on $\mu_{p^n}$. All calculations done, the natural action of $G_n$ on $<a>_n$ is via $(\kappa_n)^{-1}.(\chi_n)^{-1}$. But $a\in K^*$, so that $\chi_n =(\kappa_n)^{-1}$. Coming back to our normality condition, we thus conclude that it is equivalent to $a$ mod ${{K_n}^*}^{p^n}$ being an eigenvector for the eigencharacter $(\kappa_n)^{-1}$. Since $a\in K^*$, this is possible only for values of $n$ s.t. $K$ contains $\mu_{p^n}$.