Is the following proposition true? If yes, how would you prove this?
Proposition Let $k$ be an algebraic number field. Let $K$ be a finite abelian extension of $k$. Suppose every principal prime ideal of $k$ splits completely in $K$. Let $L$ be a finite extension of $k$. Let $E = KL$. Let $\mathfrak{P}$ be a prime ideal of $L$ such that the relative degree of $\mathfrak{P}$ in $L/k$ is 1 and $N_{L/k}(\mathfrak{P})$ is principal. Then $\mathfrak{P}$ splits completely in $E$.
Motivation
Let $\mathcal{I}$ be the group of fractional ideals of $L$. Let $\mathcal{P}$ be the group of principal ideals of $L$. Let $\mathcal{H}$ = {$I \in \mathcal{I}$; $N_{L/k}(I)$ is principal}. Then $\mathcal{H}$ is a subgroup of $\mathcal{I}$ such that $\mathcal{I} \supset \mathcal{H} \supset \mathcal{P}$. If the proposition is true, it satisfies the assumption of this proposition.
Effort
Let $N_{L/k}(\mathfrak{P}) = \mathfrak{p}$. Let $M = K \cap L$. Let $N_{L/M}(\mathfrak{P}) = \mathfrak{p}_M$. Then $\mathfrak{p}_M$ splits completely in $K$. Let $\mathfrak{p}_M = \mathfrak{Q_1}\cdots\mathfrak{Q_m}$, where $m = [K : M]$. Let $\mathfrak{Q_i'}$ be the ideal in $E$ generated by $\mathfrak{P}$ and $\mathfrak{Q_i}$. Then perhaps we can prove that $\mathfrak{P} = \mathfrak{Q_1'}\cdots\mathfrak{Q_m'}$ and $\mathfrak{Q_i'}$s are distinct primes. Hence $\mathfrak{P}$ splits completely in $E$.