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Theorem. Let $K$ and $L$ be Galois extensions of $\mathbf Q$. Set $F = K\cap L$.

Using $L = \mathbf Q(\zeta_d)$, we see the answer to the OP’s question if affirmative if and only if $\mathbf Q(\zeta_d)\cap K = \mathbf Q$.

(Two sufficient conditions to have $F = \mathbf Q$ are (i) $[K:\mathbf Q]$ and $[L:\mathbf Q]$ are relatively prime and (ii) the discriminants of $K$ and $L$ are relatively prime. When $L = \mathbf Q(\zeta_d)$, (ii) holds if $(d,{\rm disc}(K)) = 1$ since primes that ratify in $\mathbf Q(\zeta_d)$ must divide $d$.)

Proof. Case 1: $F = \mathbf Q$. By Galois theory, the composite field $KL$ is Galois over $\mathbf Q$ and ${\rm Gal}(KL/\mathbf Q) \cong {\rm Gal}(K/\mathbf Q) \times {\rm Gal}(L/\mathbf Q)$.

Case 2: $F \not= \mathbf Q$. Now there is a restriction on the conjugacy classes $C$ in ${\rm Gal}(L/\mathbf Q)$ such that some prime number $p$ (not just infinitely many) unramified in $L$ can have Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ while splitting completely in $K$. Such a prime $p$ splits completely in $F$, which implies $C \subset {\rm Gal}(L/F)$, and ${\rm Gal}(L/F)$ is a normal subgroup of ${\rm Gal}(L/\mathbf Q)$ since $F/\mathbf Q$ must be Galois. Since $F \not= \mathbf Q$, ${\rm Gal}(L/F)$ is a proper normal subgroup of ${\rm Gal}(L/\mathbf Q)$, so for $\sigma$ in ${\rm Gal}(L/\mathbf Q)$ that is not in ${\rm Gal}(L/F)$, there is no prime $p$ that is unramified in $L$, has Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ equal to the conjugacy class of $\sigma$, and splits completely in $K$.

Theorem. Let $K$ and $L$ be finite Galois extensions of $\mathbf Q$. Set $F = K\cap L$.

(3) A conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ is the Frobenius conjugacy class of some prime unramified in $L$ that splits completely in $K$ if and only if $C \subset {\rm Gal}(L/F)$, in which case $C$ is the Frobenius conjugacy class of infinitely many primes unramified in $L$ that split completely in $K$.

(Two sufficient conditions to have $F = \mathbf Q$ are (i) $[K:\mathbf Q]$ and $[L:\mathbf Q]$ are relatively prime and (ii) the discriminants of $K$ and $L$ are relatively prime. When $L = \mathbf Q(\zeta_d)$, (ii) holds if $(d,{\rm disc}(K)) = 1$ since primes that ratify in $\mathbf Q(\zeta_d)$ must divide $d$. Neither of these conditions is necessary.)

Remark. Using $L = \mathbf Q(\zeta_d)$, we see the answer to the OP’s question if affirmative if and only if $K \cap \mathbf Q(\zeta_d) = \mathbf Q$, and that even if $K \cap \mathbf Q(\zeta_d) \not= \mathbf Q$ we can still describe exactly which elements of the group $(\mathbf Z/d\mathbf Z)^\times$, viewed as ${\rm Gal}(\mathbf Q(\zeta_d)/\mathbf Q)$, contain a prime number that splits completely in $K$: it is the congruence classes mod $d$ that belong to ${\rm Gal}(\mathbf Q(\zeta_d)/F)$, where $F = K \cap \mathbf Q(\zeta_d)$.

Proof. (1) We assume $F = \mathbf Q$. By Galois theory, the composite field $KL$ is Galois over $\mathbf Q$ and ${\rm Gal}(KL/\mathbf Q) \cong {\rm Gal}(K/\mathbf Q) \times {\rm Gal}(L/\mathbf Q)$.

(2) We assume $F \not= \mathbf Q$. Now there is a restriction on the conjugacy classes $C$ in ${\rm Gal}(L/\mathbf Q)$ such that some prime number $p$ (not just infinitely many) unramified in $L$ can have Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ while splitting completely in $K$. Such a prime $p$ splits completely in $F$, which implies $C \subset {\rm Gal}(L/F)$, and ${\rm Gal}(L/F)$ is a normal subgroup of ${\rm Gal}(L/\mathbf Q)$ since $F/\mathbf Q$ must be Galois. Since $F \not= \mathbf Q$, ${\rm Gal}(L/F)$ is a proper normal subgroup of ${\rm Gal}(L/\mathbf Q)$, so for $\sigma$ in ${\rm Gal}(L/\mathbf Q)$ that is not in ${\rm Gal}(L/F)$, there is no prime $p$ that is unramified in $L$, has Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ equal to the conjugacy class of $\sigma$, and splits completely in $K$.

(3) We showed in the proof of (2) that if there is a prime unramified in $L$ that splits completely in $K$, then its Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ lies in ${\rm Gal}(L/F)$. Conversely, let $C$ be a conjugacy class of ${\rm Gal}(L/\mathbf Q)$ that lies in the normal subgroup ${\rm Gal}(L/F)$. Pick $\sigma \in C$, so $\sigma \in {\rm Gal}(L/F)$. By Galois theory the restriction mapping ${\rm Gal}(KL/K) \to {\rm Gal}(L/F)$ is an isomorphism, so we can lift $\sigma$ to an automorphism $\sigma'$ in ${\rm Gal}(KL/K)$. By Chebotarev there are (infinitely many) primes $p$ unramified in $KL$ whose Frobenius conjugacy class in ${\rm Gal}(KL/\mathbf Q)$ is the conjugacy class of $\sigma'$. Let's show for such $p$ that (i) the Frobenius conjugacy class of $p$ in ${\rm Gal}(L/\mathbf Q)$ is $C$ and (ii) $p$ splits completely in $K$:

(i) since $\sigma'|_{L} = \sigma$, the Frobenius conjugacy class of $p$ in ${\rm Gal}(L/\mathbf Q)$ is the conjugacy class of $\sigma$ in ${\rm Gal}(L/\mathbf Q)$, which is $C$,

(ii) since $\sigma'$ is trivial on $K$, the Frobenius conjugacy class of $p$ in ${\rm Gal}(K/\mathbf Q)$ is trivial, so $p$ splits completely in $K$.

Theorem. Let $K$ and $L$ be Galois extensions of $\mathbf Q$. Set $F = K\cap L$.

(1) If $F = \mathbf Q$, then for each conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ there are infinitely many primes that are unramified in $L$ with Frobebius conjugacy class $C$ and split completely in $K$.

(2) If $F \not= \mathbf Q$ then there is a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ such that no prime unramified in $L$ with Frobenius conjugacy class $C$ splits completely in $K$.

Using $L = \mathbf Q(\zeta_d)$, we see the answer to the OP’s question if affirmative if and only if $\mathbf Q(\zeta_d)\cap K = \mathbf Q$.

(Two sufficient conditions to have $F = \mathbf Q$ are (i) $[K:\mathbf Q]$ is relatively prime to $[L:\mathbf Q]$ and (ii) the discriminants of $K$ and $L$ are relatively prime. When $L = \mathbf Q(\zeta_d)$, (ii) holds if $(d,{\rm disc}(K)) = 1$ since primes that ratify in $\mathbf Q(\zeta_d)$ must divide $d$.)

Proof.

Case 1: $F = \mathbf Q$. By Galois theory, the composite field $KL$ is Galois over $\mathbf Q$ and ${\rm Gal}(KL/\mathbf Q) \cong {\rm Gal}(K/\mathbf Q) \times {\rm Gal}(L/\mathbf Q)$.

Pick a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$. Then $\{1\} \times C$ is a conjugacy class in ${\rm Gal}(KL/\mathbf Q)$. By Chebotarev, there are infinitely many primes $p$ unramified in $KL$ such that its Frobenius conjugacy class in ${\rm Gal}(KL/\mathbf Q)$ is $\{1\} \times C$, so such $p$ split completely in $K$ while having Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$.

Case 2: $F \not= \mathbf Q$. Now there is a restriction on the conjugacy classes $C$ in ${\rm Gal}(L/\mathbf Q)$ such that some prime number $p$ (not just infinitely many) unramified in $L$ can have Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ while splitting completely in $K$. Such a prime $p$ splits completely in $F$, which implies $C \subset {\rm Gal}(L/F)$, a normal subgroup of ${\rm Gal}(L/\mathbf Q)$ since $F/\mathbf Q$ must be Galois. Since $F \not= \mathbf Q$, ${\rm Gal}(L/\mathbf Q)$ is a proper normal subgroup of ${\rm Gal}(L/\mathbf Q)$, so for $\sigma$ in ${\rm Gal}(L/\mathbf Q)$ that is not in ${\rm Gal}(L/F)$, there is no prime $p$ that is unramified in $L$, has Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ equal to the conjugacy class of $\sigma$, and splits completely in $K$.

QED

Theorem. Let $K$ and $L$ be Galois extensions of $\mathbf Q$. Set $F = K\cap L$.

(1) If $F = \mathbf Q$, then for each conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ there are infinitely many primes that are unramified in $L$ with Frobebius conjugacy class $C$ and split completely in $K$.

(2) If $F \not= \mathbf Q$ then there is a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ such that no prime unramified in $L$ with Frobenius conjugacy class $C$ splits completely in $K$.

Using $L = \mathbf Q(\zeta_d)$, we see the answer to the OP’s question if affirmative if and only if $\mathbf Q(\zeta_d)\cap K = \mathbf Q$.

(Two sufficient conditions to have $F = \mathbf Q$ are (i) $[K:\mathbf Q]$ and $[L:\mathbf Q]$ are relatively prime and (ii) the discriminants of $K$ and $L$ are relatively prime. When $L = \mathbf Q(\zeta_d)$, (ii) holds if $(d,{\rm disc}(K)) = 1$ since primes that ratify in $\mathbf Q(\zeta_d)$ must divide $d$.)

Proof. Case 1: $F = \mathbf Q$. By Galois theory, the composite field $KL$ is Galois over $\mathbf Q$ and ${\rm Gal}(KL/\mathbf Q) \cong {\rm Gal}(K/\mathbf Q) \times {\rm Gal}(L/\mathbf Q)$.

Pick a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$. Then $\{1\} \times C$ is a conjugacy class in ${\rm Gal}(KL/\mathbf Q)$. By Chebotarev, there are infinitely many primes $p$ unramified in $KL$ such that its Frobenius conjugacy class in ${\rm Gal}(KL/\mathbf Q)$ is $\{1\} \times C$, so such $p$ split completely in $K$ while having Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$.

Case 2: $F \not= \mathbf Q$. Now there is a restriction on the conjugacy classes $C$ in ${\rm Gal}(L/\mathbf Q)$ such that some prime number $p$ (not just infinitely many) unramified in $L$ can have Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ while splitting completely in $K$. Such a prime $p$ splits completely in $F$, which implies $C \subset {\rm Gal}(L/F)$, and ${\rm Gal}(L/F)$ is a normal subgroup of ${\rm Gal}(L/\mathbf Q)$ since $F/\mathbf Q$ must be Galois. Since $F \not= \mathbf Q$, ${\rm Gal}(L/F)$ is a proper normal subgroup of ${\rm Gal}(L/\mathbf Q)$, so for $\sigma$ in ${\rm Gal}(L/\mathbf Q)$ that is not in ${\rm Gal}(L/F)$, there is no prime $p$ that is unramified in $L$, has Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ equal to the conjugacy class of $\sigma$, and splits completely in $K$.

QED

Let $L$ be a Galois extension of $\mathbf Q$ such that $K \cap L = \mathbf Q$. (For example, this holds if (i) $[L:\mathbf Q]$ is relatively prime to $[K:\mathbf Q]$ or (ii) the discriminant of $L$ is relatively prime to the discriminant of $K$.) Then the composite field $KL$ is Galois over $\mathbf Q$ and ${\rm Gal}(KL/\mathbf Q) \cong {\rm Gal}(K/\mathbf Q) \times {\rm Gal}(L/\mathbf Q)$.

Pick a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$. Then $\{1\} \times C$ is a conjugacy class in ${\rm Gal}(KL/\mathbf Q)$. By Chebotarev, there are infinitely many primes $p$ unramified in $KL$ such that its Frobenius conjugacy class in ${\rm Gal}(KL/\mathbf Q)$ is $\{1\} \times C$, so such $p$ split completely in $K$ while having Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$.

When $L = \mathbf Q(\zeta_d)$, it fits the condition $K \cap L = \mathbf Q$ if $(\varphi(d),[K:\mathbf Q]) = 1$ or if $(d,{\rm disc}(K)) = 1$; in the second case the discriminants of $K$ and $L$ are relatively prime since a prime number that ramifies in $L$ must divide $d$. Since ${\rm Gal}(L/\mathbf Q) \cong (\mathbf Z/d\mathbf Z)^\times$, a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ is just an arithmetic progression $a \bmod d$ where $\gcd(a,d) = 1$. When $p \nmid d$ the Frobenius at $p$ in ${\rm Gal}(L/\mathbf Q)$ is $p \bmod d$, so there are infinitely many primes $p$ such that $p \equiv a \bmod d$ and $p$ splits completely in $K$.

Returning to the case of general $L$ Galois over $\mathbf Q$, if the intersection $F = K \cap L$ is bigger than $\mathbf Q$ then there is a restriction on the conjugacy classes $C$ in ${\rm Gal}(L/\mathbf Q)$ such that some prime number $p$ (not just infinitely many) unramified in $L$ can have Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ while splitting completely in $K$. Such $p$ splits completely in $F$, which implies $C \subset {\rm Gal}(L/F)$, a normal subgroup of ${\rm Gal}(L/\mathbf Q)$ since $F/\mathbf Q$ must be Galois. If $F \not= \mathbf Q$ then ${\rm Gal}(L/\mathbf Q)$ is a proper normal subgroup of ${\rm Gal}(L/\mathbf Q)$, so for $\sigma$ in ${\rm Gal}(L/\mathbf Q)$ that does not fix $F$ pointwise there is no prime $p$ unramified in $L$ that splits completely in $K$.

For example, if $F = \mathbf Q(\zeta_d) \cap K$ is bigger than $\mathbf Q$ and the proper subgroup ${\rm Gal}(\mathbf Q(\zeta_d)/F)$ of ${\rm Gal}(\mathbf Q(\zeta_d)/\mathbf Q) \cong (\mathbf Z/d\mathbf Z)^\times$ does not contain the arithmetic progression $b \bmod d$ where $(b,d) = 1$, none of the infinitely many primes $p \equiv b \bmod d$ split completely in $K$.

Theorem. Let $K$ and $L$ be Galois extensions of $\mathbf Q$. Set $F = K\cap L$.

(1) If $F = \mathbf Q$, then for each conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ there are infinitely many primes that are unramified in $L$ with Frobebius conjugacy class $C$ and split completely in $K$.

(2) If $F \not= \mathbf Q$ then there is a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ such that no prime unramified in $L$ with Frobenius conjugacy class $C$ splits completely in $K$.

Using $L = \mathbf Q(\zeta_d)$, we see the answer to the OP’s question if affirmative if and only if $\mathbf Q(\zeta_d)\cap K = \mathbf Q$.

(Two sufficient conditions to have $F = \mathbf Q$ are (i) $[K:\mathbf Q]$ is relatively prime to $[L:\mathbf Q]$ and (ii) the discriminants of $K$ and $L$ are relatively prime. When $L = \mathbf Q(\zeta_d)$, (ii) holds if $(d,{\rm disc}(K)) = 1$ since primes that ratify in $\mathbf Q(\zeta_d)$ must divide $d$.)

Proof.

Case 1: $F = \mathbf Q$. By Galois theory, the composite field $KL$ is Galois over $\mathbf Q$ and ${\rm Gal}(KL/\mathbf Q) \cong {\rm Gal}(K/\mathbf Q) \times {\rm Gal}(L/\mathbf Q)$.

Pick a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$. Then $\{1\} \times C$ is a conjugacy class in ${\rm Gal}(KL/\mathbf Q)$. By Chebotarev, there are infinitely many primes $p$ unramified in $KL$ such that its Frobenius conjugacy class in ${\rm Gal}(KL/\mathbf Q)$ is $\{1\} \times C$, so such $p$ split completely in $K$ while having Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$.

Case 2: $F \not= \mathbf Q$. Now there is a restriction on the conjugacy classes $C$ in ${\rm Gal}(L/\mathbf Q)$ such that some prime number $p$ (not just infinitely many) unramified in $L$ can have Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ while splitting completely in $K$. Such a prime $p$ splits completely in $F$, which implies $C \subset {\rm Gal}(L/F)$, a normal subgroup of ${\rm Gal}(L/\mathbf Q)$ since $F/\mathbf Q$ must be Galois. Since $F \not= \mathbf Q$, ${\rm Gal}(L/\mathbf Q)$ is a proper normal subgroup of ${\rm Gal}(L/\mathbf Q)$, so for $\sigma$ in ${\rm Gal}(L/\mathbf Q)$ that is not in ${\rm Gal}(L/F)$, there is no prime $p$ that is unramified in $L$, has Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ equal to the conjugacy class of $\sigma$, and splits completely in $K$.

QED