nt.number theory - On periods of algebraic integers modulo rational primes

I run, somewhat indirectly, into the following problem and I have no hints where to look in the literature in search for answers or clues.

Let $K$ be a number field, which we may assume Galois if it helps, $cal O$ its ring of integers and for each prime number $p$ let $R_p={cal O}/p{cal O}$ (a finite product of finite fields of characteristic $p$ for almost all $p$). Fix $lambdain{cal O}$, $lambdaneq0$ or a root of $1$. Then $barlambdain R_p^times$ for almost all $p$ and the period $pi_p=pi_p(lambda)$ of $lambda$ is defined as the smallest positive $d$ such that $barlambda^d=1$.

It is obvious that if we fix an integer $n$ the number of $p$'s such that $pi_pleq n$ is finite, since $pi_p=d$ implies that $p|(lambda^d-1)$ and there are only finitely many of those.

On the other hand, if $ngeq2$ the number of $p$'s such that
$max{text{Supp}(pi_p)}leq n$ (the support $text{Supp}(N)$ of an integer $N$ is the set of prime divisors of $N$) is infinite. For instance, the set of elements $lambda^{2^k}-1$ has an infinite set of rational prime divisors because $lambda^{2^{k+1}}-1=(lambda^{2^k}-1)(lambda^{2^k}+1)$ and the only common prime divisors in $cal O$ to the 2 factors are primes of residual characteristic 2. Thus, as k grows, a new set of primes adds up at each step, so to speak.

Now the question is: fix an arithmetic progression ${cal P}:a,a+d,a+2d,dots$ with $(a,d)=1$, is it true that there are infinitely many primes in $cal P$ such that $max{text{Supp}(pi_p)}leq n$? Conditionally on $n$?

In particular: suppose $K$ quadratic, and let $ell>2$ a prime. Are there infinitely many primes $pequiv 1bmodell$ such that $max{text{Supp}(pi_p)}leqell-1$?