Do we have such a direct product decomposition of Galois groups?

You can write your $L$ as the compositum of $\mathbb{Q}(\zeta_{p^k})$ and $\mathbb{Q}(\zeta_{n})$. Since $(p,n)=1$, the two are disjoint over $\mathbb{Q}$, and so the Galois group of $L$ is isomorphic to the direct product of the two Galois groups, one of which is $E$. Let's call the other subgroup $H$. Now, every element $g$ of $G$ is uniquely a product of an element $\epsilon$ of $E$ and an element $h$ of $H$. $E$ is contained in $D$, so $\epsilon h$ fixes $P$ if and only if $h$ does. In other words, the decomposition group $D$ is generated by $E$ and the decomposition group of $P$ in ${\rm Gal}(\mathbb{Q}(\zeta_{p^kn})/\mathbb{Q}(\zeta_{p^k}))=H$, so is indeed a direct product.