Proof.
This is mostly a reformulation of the results on finite Galois extensions proved in More on Algebra, Section 15.112. The surjectivity of the map $D \to \text{Aut}(\kappa (\mathfrak m)/\kappa )$ is More on Algebra, Lemma 15.110.10. This gives the first exact sequence.
To construct the second short exact sequence let $\Lambda $ be the set of finite Galois subextensions, i.e., $\lambda \in \Lambda $ corresponds to $L/L_\lambda /K$. Set $G_\lambda = \text{Gal}(L_\lambda /K)$. Recall that $G_\lambda $ is an inverse system of finite groups with surjective transition maps and that $G = \mathop{\mathrm{lim}}\nolimits _{\lambda \in \Lambda } G_\lambda $, see Fields, Lemma 9.22.3. We let $B_\lambda $ be the integral closure of $A$ in $L_\lambda $. Then we set $\mathfrak m_\lambda = \mathfrak m \cap B_\lambda $ and we denote $P_\lambda , I_\lambda , D_\lambda $ the wild inertia, inertia, and decomposition group of $\mathfrak m_\lambda $, see More on Algebra, Lemma 15.112.5. For $\lambda \geq \lambda '$ the restriction defines a commutative diagram
\[ \xymatrix{ P_\lambda \ar[d] \ar[r] & I_\lambda \ar[d] \ar[r] & D_\lambda \ar[d] \ar[r] & G_\lambda \ar[d] \\ P_{\lambda '} \ar[r] & I_{\lambda '} \ar[r] & D_{\lambda '} \ar[r] & G_{\lambda '} } \]
with surjective vertical maps, see More on Algebra, Lemma 15.112.10.
From the definitions it follows immediately that $I = \mathop{\mathrm{lim}}\nolimits I_\lambda $ and $D = \mathop{\mathrm{lim}}\nolimits D_\lambda $ under the isomorphism $G = \mathop{\mathrm{lim}}\nolimits G_\lambda $ above. Since $L = \mathop{\mathrm{colim}}\nolimits L_\lambda $ we have $B = \mathop{\mathrm{colim}}\nolimits B_\lambda $ and $\kappa (\mathfrak m) = \mathop{\mathrm{colim}}\nolimits \kappa (\mathfrak m_\lambda )$. Since the transition maps of the system $D_\lambda $ are compatible with the maps $D_\lambda \to \text{Aut}(\kappa (\mathfrak m_\lambda )/\kappa )$ (see More on Algebra, Lemma 15.112.10) we see that the map $D \to \text{Aut}(\kappa (\mathfrak m)/\kappa )$ is the limit of the maps $D_\lambda \to \text{Aut}(\kappa (\mathfrak m_\lambda )/\kappa )$.
There exist canonical maps
\[ \theta _{\lambda , can} : I_\lambda \longrightarrow \mu _{n_\lambda }(\kappa (\mathfrak m_\lambda )) \]
where $n_\lambda = |I_\lambda |/|P_\lambda |$, where $\mu _{n_\lambda }(\kappa (\mathfrak m_\lambda ))$ has order $n_\lambda $, such that $\theta _{\lambda , can}(\tau \sigma \tau ^{-1}) = \tau (\theta _{\lambda , can}(\sigma ))$ for $\tau \in D_\lambda $ and $\sigma \in I_\lambda $, and such that we get commutative diagrams
\[ \xymatrix{ I_\lambda \ar[r]_-{\theta _{\lambda , can}} \ar[d] & \mu _{n_\lambda }(\kappa (\mathfrak m_\lambda )) \ar[d]^{(-)^{n_\lambda /n_{\lambda '}}} \\ I_{\lambda '} \ar[r]^-{\theta _{\lambda ', can}} & \mu _{n_{\lambda '}}(\kappa (\mathfrak m_{\lambda '})) } \]
see More on Algebra, Remark 15.112.11.
Let $S \subset \mathbf{N}$ be the collection of integers $n_\lambda $. Since $\Lambda $ is directed, we see that $S$ is multiplicatively directed. By the displayed commutative diagrams above we can take the limits of the maps $\theta _{\lambda , can}$ to obtain
\[ \theta _{can} : I \to \mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\kappa (\mathfrak m)). \]
This map is continuous (small detail omitted). Since the transition maps of the system of $I_\lambda $ are surjective and $\Lambda $ is directed, the projections $I \to I_\lambda $ are surjective. For every $\lambda $ the diagram
\[ \xymatrix{ I \ar[d] \ar[r]_-{\theta _{can}} & \mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\kappa (\mathfrak m)) \ar[d] \\ I_{\lambda } \ar[r]^-{\theta _{\lambda , can}} & \mu _{n_\lambda }(\kappa (\mathfrak m_\lambda )) } \]
commutes. Hence the image of $\theta _{can}$ surjects onto the finite group $\mu _{n_\lambda }(\kappa (\mathfrak m)) = \mu _{n_\lambda }(\kappa (\mathfrak m_\lambda ))$ of order $n_\lambda $ (see above). It follows that the image of $\theta _{can}$ is dense. On the other hand $\theta _{can}$ is continuous and the source is a profinite group. Hence $\theta _{can}$ is surjective by a topological argument.
The property $\theta _{can}(\tau \sigma \tau ^{-1}) = \tau (\theta _{can}(\sigma ))$ for $\tau \in D$, $\sigma \in I$ follows from the corresponding properties of the maps $\theta _{\lambda , can}$ and the compatibility of the map $D \to \text{Aut}(\kappa (\mathfrak m))$ with the maps $D_\lambda \to \text{Aut}(\kappa (\mathfrak m_\lambda ))$. Setting $P = \mathop{\mathrm{Ker}}(\theta _{can})$ this implies that $P$ is a normal subgroup of $D$. Setting $I_ t = I/P$ we obtain the isomorphism $I_ t \to \mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\kappa (\mathfrak m))$ from the surjectivity of $\theta _{can}$.
To finish the proof we show that $P = \mathop{\mathrm{lim}}\nolimits P_\lambda $ which proves that $P$ is a pro-$p$-group. Recall that the tame inertia group $I_{\lambda , t} = I_\lambda /P_\lambda $ has order $n_\lambda $. Since the transition maps $P_\lambda \to P_{\lambda '}$ are surjective and $\Lambda $ is directed, we obtain a short exact sequence
\[ 1 \to \mathop{\mathrm{lim}}\nolimits P_\lambda \to I \to \mathop{\mathrm{lim}}\nolimits I_{\lambda , t} \to 1 \]
(details omitted). Since for each $\lambda $ the map $\theta _{\lambda , can}$ induces an isomorphism $I_{\lambda , t} \cong \mu _{n_\lambda }(\kappa (\mathfrak m))$ the desired result follows.
$\square$
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