The idea isn’t too far off, but you need to note that $[-]$ is far from being surjective, so it won’t map norm groups to norm groups… I really hope I got this right.
Clearly, if $L/K$ is finite abelian, any completion of $L$ at a place above $\mathfrak{p}$ is abelian over $K_{\mathfrak{p}}$. So what remains to be proved is that, for any finite abelian $L/K_{\mathfrak{p}}$, there exists an abelian extension $E/K$ such that $L \subset EK_{\mathfrak{p}}$.
Passing to norm subgroups, and using (5.8), what needs to be shown is that for every norm subgroup $N \leq K_{\mathfrak{p}}^{\times}$ (corresponding to $L$) there exists a finite index open subgroup $N’ \leq C_K$ such that $[-]^{-1}(N’) \subset N$. In other words, we have to show that there exists a finite index open subgroup $K^{\times} \leq N’ \leq \mathbb{A}_K^{\times}$ such that $[-]^{-1}(N’) \subset N$.
Assume $N \supset N_1 \cap N_2$ where we know the solution for $N_1$ and $N_2$ – ie we know satisfactory $N’_1$ and $N’_2$. Then $N’=N’_1 \cap N’_2$ clearly works for $N$.
Let $\pi \in K$ be such that $(\pi)$ is a nontrivial power of $\mathfrak{p}$. We know that $N$ contains $\{u\pi^n,\, u \in 1+\mathfrak{p}^{\alpha}, \beta|n\}$ for $\alpha >0$, $\beta >0$. Let $N_1=\pi^{\mathbb{Z}}(1+\mathfrak{p}^{\alpha})$, $N_2=\mathcal{O}_{K_{\mathfrak{p}}}^{\times}\pi^{\beta \mathbb{Z}}$. Then $N \supset N_1 \cap N_2$.
$N_2$ is the norm group of the non-ramified extension $K_{\mathfrak{p}}(\mu_{q^{\nu\beta}-1})$ (where $\pi$ has valuation $\nu$, the order of $\mathfrak{p}$ in the class group and $q$ is the cardinal of the residue field at $\mathfrak{p}$) which clearly comes from a global extension (with the associated $N’_2$). So it’s enough to solve the case $N=N_1$.
In other words, we want to construct a finite index open subgroup $N’$ of $\mathbb{A}_K^{\times}$ containing $K^{\times}$ such that its only elements with support in $K_{\mathfrak{p}}$ are in $\pi^{\mathbb{Z}}(1+\mathfrak{p}^{\alpha})$.
Let $S$ be a finite set of finite places of $K$ containing $\mathfrak{p}$, $S’$ the other finite places, $S_{\infty}$ the infinite places. We will take $N’=K^{\times}F$, where $F=\prod_{v \in S_{\infty}}{K_v^{\times}} \prod_{v \in S’}{\mathcal{O}_{K_v}^{\times}}\prod_{v \in S}{H_v}$ where $H_v \subset \mathcal{O}_{K_v}^{\times}$, which is an open subgroup (then $N’$ is open and finite index essentially because the $\mathcal{O}_v^{\times}$ are compact and the class group is finite).
To show the statement, we need to see that $H_{\mathfrak{p}} \subset (1+\mathfrak{p}^{\alpha})$ and if $x \in K^{\times}$ is in $H_v$ for all $v \in S_0 :=S \backslash \{\mathfrak{p}\}$, then $x \in \pi^{\mathbb{Z}}(1+\mathfrak{p})^{\alpha})$.
Now by the assumption on $\pi$, $\mathcal{O}_K^{\mathfrak{p},\times}$ (the superscript means that we require invertibility everywhere but at $\mathfrak{p}$) is the direct product $\pi^{\mathbb{Z}}\times \mathcal{O}_K^{\times}$. So we want to show that if $x\in \pi^{\mathbb{Z}}\times \mathcal{O}_K^{\times} $ satisfies a certain “non-$\mathfrak{p}$-congruence then $x \in \pi^{\mathbb{Z}}(1+\mathfrak{p}^{\alpha})$.
It is enough to show that, for some integer $f$ (the index of the right subgroup), a non-$\mathfrak{p}$ congruence condition can cut out the set $fG$ of $f$-powers in $G=\pi^{\mathbb{Z}} \times \mathcal{O}_K^{\times}$.
Now, let $P$ be the product of all the unit groups of the residual fields at finite places not $\mathfrak{p}$. By (*) below, $G \rightarrow P$ is injective with a cokernel without $f$-torsion, ie $G/fG \rightarrow P/fP$ is injective, and thus ($G$ has finite rank) there is an injection $G/fG$ to $P’/fP’$ where $P’$ is the product of finitely many multiplicative groups of residue fields at $v\neq\mathfrak{p}$ — and this concludes.
Lemma(*): Let $S$ be a cofinite set of finite places of a number field $K$ and $x \in K^{\times}, f \geq 1$. Assume that $x$ is a $f$-power in every residue field at a place of $S$ – then $x$ is a $f$-th power in $K$.
Proof: We can always assume that $S$ contains only places coprime to $f$ and $x$.
Let $L=K(e^{2i\pi/f},x^{1/f})$. Then $L/K$ is Galois. Each place $w$ of $L$ above some $v \in S$ has a well-defined Frobenius in $Gal(L/K)$ acting on the residue field at $w$, and it fixes one of the roots mod $w$ of the minimal polynomial $\mu$ of $x^{1/f}$. As $L/K$ is unramified at $w$, it means that this Frobenius (acting on the global roots of $\mu$) has a fixed point. By Cebotarev, every element of $Gal(L/K)$ fixes one of the roots of $\mu$. But $Gal(L/K)$ acts transitively on them – and by Burnside lemma it follows that $\mu$ has only a root and $L=K$.
Best Answer
As pointed out in the comments, we have $N_{M/K} = N_{L/K} \circ N_{M/L},$ so that $$N_{M/K}(C_M) = N_{L/K}(N_{M/L}(C_M)) \subset N_{L/K}(C_L).$$ This yields a quotient morphism $$C_K/N_{M/K}(C_M) \to C_K/N_{L/K}(C_L).$$