The answer to 1 is negative, but 2 is true (see below).

Here is a counterexample to 1 (as well as to finiteness of $L \to M$).

**Example.** Let $K = \mathbf Q(i)$ and let $L = K\big(\pmb\mu_{2^\infty},\sqrt[2^\infty]{2+i}\big)$ be the smallest Galois extension containing all $2^n$-th roots of $2+i$ for all $n \geq 1$ (where $\pmb\mu_m \subseteq \bar{\mathbf Q}$ denotes the $m$-th roots of unity). Note that $K \to L$ is Galois with group
\begin{align*}
\operatorname{Gal}(L/K) \stackrel\sim\to&\ \mathbf{Aff}(\mathbf Z_2) = \left\{\begin{pmatrix}a & b \\ 0 & 1\end{pmatrix} \in \operatorname{GL}_2(\mathbf Z_2)\ \Bigg|\ a \in \mathbf Z_2^\times, b \in \mathbf Z_2\right\},\\
\left(\begin{array}{rl}\zeta_{2^n}\!\!\!\! & \mapsto \zeta_{2^n}^a \\ \sqrt[2^n]{2+i}\!\!\!\! & \mapsto \zeta_{2^n}^b\sqrt[2^n]{2+i} \end{array}\right) \mathrel{\unicode{x21a4}}&\ \!\begin{pmatrix}a & b \\ 0 & 1\end{pmatrix}.
\end{align*}
(This isomorphism probably depends on a choice of compatible primitive $2^n$-th roots of unity $\zeta_{2^n}$, in the sense that that $\zeta_{2^n}^{2^m} = \zeta_{2^{n-m}}$ for all $n \geq m \geq 0$. There is probably a version possible with $b \in \mathbf Z_2(1)$ if you don't want to make such a choice.)

Then the Galois closure $M$ of $\mathbf Q \to L$ is given by further adjoining $\sqrt[2^n]{2-i}$ for all $n$. This has infinite ramification index over the prime $(2-i) \subseteq \mathcal O_K$. Since $(2-i)$ is unramified in $L$, that means that $L \to M$ still has infinite ramification index over any prime above $(2-i)$. $\square$

**Remark.** The answer to 2 is positive, and in fact $\operatorname{Gal}(M/\mathbf Q)$ is a closed subgroup of $\operatorname{GL}_{d \cdot [K:\mathbf Q]}(\mathbf Z_p)$. Indeed, if $G = \operatorname{Gal}(\bar{\mathbf Q}/\mathbf Q)$ and $H = \operatorname{Gal}(\bar{\mathbf Q}/K)$, then $H \subseteq G$ is a finite index closed subgroup (i.e. open subgroup), and by assumption we have a map
$$\rho \colon H \to \operatorname{GL}_d(\mathbf Z_p)$$
whose kernel is $\operatorname{Gal}(\bar{\mathbf Q}/L)$. Viewing $\rho$ as a representation of $H$ on the module $A = \mathbf Z_p^d$ gives an *induced representation* (some authors call this the *coinduced representation*)
$$B := \operatorname{Ind}_G^H(A) = \left\{\phi \in \operatorname{Map}_{\text{cts}}(G,A)\ \big|\ \phi(hg) = h\phi(g) \text{ for all } g \in G, h \in H\right\},$$
where $g \in G$ acts on $\phi \colon G \to A$ via $(g\phi)(\sigma) = \phi(\sigma g)$. For $a \in A$ and $\sigma \in G$, note that the function $\phi_{a,\sigma} \colon G \to A$ given by
$$g \mapsto \begin{cases} (g\sigma^{-1})a, & g\sigma^{-1} \in H, \\ 0, & \text{else}. \end{cases}$$
is in $B$. Since $[G:H] = [K:\mathbf Q]$ is finite, we see that $B$ is again a finite free $\mathbf Z_p$-module. Write $\tau \colon G \to \operatorname{Aut}(B)$ for the corresponding group homomorphism. Here is a well-known lemma:

**Lemma.** *Let $H \subseteq G$ be an open subgroup of a profinite group $G$, let $\rho \colon H \to \operatorname{Aut}(A)$ be a continuous representation on a (topologically) finitely generated profinite abelian group $A$, and let $\tau \colon G \to \operatorname{Aut}(B)$ be its induced representation. Then $\tau$ is continuous, and $\ker(\tau) = \bigcap_{\sigma \in G} \sigma\ker(\rho)\sigma^{-1}$.*

*Proof.* For the second statement, write $U = \ker(\rho)$. Then $g \in \ker(\tau)$ means that for all $\phi \in \operatorname{Ind}_G^H(A)$ and all $\sigma \in G$, we have $\phi(\sigma g) = \phi(\sigma)$. Applying this to $\phi_{a,\sigma}$ for $a \in A \setminus \{0\}$ shows that $\sigma g \sigma^{-1}$ is in $H$ and fixes $a$. Running over all $a$ shows that $\sigma g \sigma^{-1} \in U$, so $g \in \bigcap_{\sigma \in G} \sigma U \sigma^{-1}$ since $\sigma \in G$ was arbitrary. Conversely, if $g \in \bigcap_{\sigma \in G} \sigma U \sigma^{-1}$ and $\sigma \in G$, then $g = \sigma^{-1}h\sigma$ for $h \in U$, so $\sigma g = h\sigma$, hence $\phi(\sigma g) = \phi(h \sigma) = h\phi(\sigma) = \phi(\sigma)$ for all $\phi \in \operatorname{Ind}_G^H(A)$. This proves the second claim.

For the first statement, note that the subgroups $G_n := \ker(\operatorname{Aut}(B) \twoheadrightarrow \operatorname{Aut}(B/nB))$ form a basis of open neighbourhoods of $1$ in $\operatorname{Aut}(B)$, so it suffices to show that $G_n$ is open for all $n \geq 1$. Since $\operatorname{Ind}_G^H(A/nA) = B/nB$, applying the first statement to $A/nA$ shows that
$$G_n = \bigcap_{\sigma \in G} \sigma \ker\big(H \to \operatorname{Aut}(A/nA)\big) \sigma^{-1}.$$
Since $\ker(H \to \operatorname{Aut}(A/nA))$ is open by assumption and this intersection can be carried out over a (finite) set of coset representatives for $G/H$, we conclude that $G_n$ is open. $\square$

Thus, the induced representation $\operatorname{Gal}(\bar{\mathbf Q}/\mathbf Q) \to \operatorname{Aut}(B) = \operatorname{GL}_{d \cdot [K:\mathbf Q]}(\mathbf Z_p)$ is a continuous homomorphism whose kernel corresponds to the Galois closure $\mathbf Q \to M$ of $\mathbf Q \to L$, hence gives a continuous injection
$$\operatorname{Gal}(M/\mathbf Q) \hookrightarrow \operatorname{GL}_m(\mathbf Z_p).$$
Such a map is automatically closed since everything is profinite, and $\operatorname{Gal}(M/L)$ is in turn a closed subgroup of $\operatorname{Gal}(M/\mathbf Q)$. $\square$