This question is based on Theorem 10.2 of John Howie's 'Fields and Galois Theory', p.171. All fields have characteristic zero.
$M$ is a normal radical field extension of $K$, i.e. $M=K(\alpha_1,\alpha_2,…,\alpha_n)$ where $\alpha_i^{p_i}\in K(\alpha_1,…\alpha_{i-1})$ for some prime $p_i$.
We also define $P=M(\omega)$ where $\omega$ is a primitive $p_i$th root of unity. Howie points out that, as $P$ is a splitting field for $X^{p_i}-1$ over $M$, $P/M$ is a normal extension of $M$. So far so good.
As we have $M/K$ normal then, using the Fundamental Theorem of Galois Theory (FTGT), we have $\text{Gal}(P/M)\vartriangleleft \text{Gal}(P/K)$. Also by FTGT we have the following isomorphism:
$$\text{Gal}(M/K)\simeq\frac{\text{Gal}(P/K)}{\text{Gal}(P/M)}$$
However, in order for the FTGT to apply, I understand that we also require $P$ to be a normal extension of $K$. Howie does not seem to be explicit about why this condition holds. The only strategy I know would be to show that $P=K(\omega,\alpha_1,\alpha_2,…,\alpha_n)$ is a splitting field for a polynomial over $K$, however it is not obvious to me what that polynomial would be, or how I could show that its splitting fields was the one required.
Best Answer
You assume that $M/K$ is a normal extension. This means there is a polynomial $f\in K[x]$ such that $M$ is a splitting field of $f$ over $K$. But then note that $P$ is a splitting field of $(x^{p_i}-1)f(x)\in K[x]$, and hence $P/K$ is normal.
Note that in general if $P/M$ and $M/K$ are both normal extensions it doesn't imply that $P/K$ must be normal. Just in your case we are lucky that $x^{p_i}-1$ is also a polynomial in $K[x]$.