Let $N$ be a positive integer and $p$ a prime not dividing $N$.
Let $K$ be a finite Galois extension of $\mathbb{Q}$ which is unramified outside the primes dividing $Np$. Let $G$ be the Galois group of $K$ over $\mathbb{Q}$. Then for any conjugacy class $C \subset G$, there is a prime number $q$ not dividing $Np$ such that $\text{Frob}\_q = C$. 


My question is the following: In the same situation as above, for any conjugate $C \subset G$ does there always exist a prime number $q$ not dividing $Np$ such that


(a) $\text{Frob}\_q=C$;


(b) $q\not\equiv 1 \pmod p$?



a variant of Cebotarev density theorem
As noted in the comments, this is false in general. For instance, if $K=\mathbb Q(e^{2i \pi/p})$, then $G=(\mathbb Z/p\mathbb Z)^\ast$. If you take $C=\{1\}$, then condition (a) is that $q \equiv 1 \pmod{p}$ and condition (b) is that $q \not \equiv 1 \pmod{p}$ which makes finding such primes $q$ a desperate task.


There is however a general situation where the answer is positive: when the group $G$ has no quotient isomorphic to a quotient of $(\mathbb Z/p \mathbb Z)^\ast$, in particular, when the group $G$ is perfect (equal to its own derived subgroup). In this case, indeed, the extension $K(e^{2i \pi/p})/\mathbb Q$ has a Galois groups with subjects to both $G$ and $(\mathbb Z/p\mathbb Z)^\ast$, hence which is isomorphic by Goursat's lemma and our assumption to $G \times (\mathbb Z/p\mathbb Z)^\ast$,
and by Cebotarev you can find primes $q$ such that $Frob\_q$ is in the class $C \times 1$ in this group. (Of course, this is a very standard argument in the field)


