The problem is completely incomprehensible. What is "such for their there is (a,b,c,q) with this terms is finite " supposed to mean? Can anyone give a clear explanation of the problem?

It means, prove that, there are infinitely many primes p such that there are only finite number of (a,b,c,q) which provide i) and ii) conditions hold.

@above but i think it means that for every constant a,b,c,q there are finitely many primes that those two conditions hold. which is obvious if im not wrong ....

I think q can depend on p.the problem is this: for constant a,b,c,d set of nautral numbers (p,q) p is prime satisfying those condition is finite. proof:if $A\equiv B$ we have $\{q^n\}=\{a/c,a/c+b/c,...,a/c+db/c\}$ by primitive root we havr the set $\{q^n\}$ is the set of the roots $x^d-1=0 modp$ so we must have $x^d-1=(x-a)(x-(a+b)/c)...(x-(a+db)/c)$ but it is easy to see if two rational polynomials equal for all primes p they are equal.

The original statement: $2 \leq d$ is a natural number. $B_{a,b}=\{a,a+b,a+2b,...,a+db\}$ $A_{c,q}=\{cq^n \vert n \in\mathbb{N}\}$ Prove that there are only finitely many primes $p$ for which there exist a quadruple $(a,b,c,q)$ of positive integers satisfying: $i$ ) $ p \nmid abcq $ $ ii$ ) $A_{c,q} \equiv B_{a,b} \mod p $

Replace $d$ by $d-1$ in the problem so we actually have $d \ge 3$ being a natural number, $B_{a,b} = \{a,a+b,...,a+(d-1)b\}$. Consider $p \gg d$. Divide everything by $c$ it suffices to consider the case $c = 1$. We can prove the following in order: $\delta_p(q) = d$ and $d | p-1$. Moreover $\{cq^n|n = 1,...,p-1\} = \{a, a+b,...,a+db\}$ up to multiplicity. Sum over $1$st powers of elements of $A_{c,q}$ to get $p|2a + (d-1)b$. Sum over $2$nd powers of elements of $A_{c,q}$ combining with the above expression to get $p|b-a$. Rewrite the divisibility from summing 1st powers to get $p|d+1$ This means that $d = p-1$, contradiction to $p \gg d$.