Looking at a problem in representation theory I ran into a question on small primes in arithmetic progressions.

Let me begin with a short summary of results on small primes in arithmetic progressions. By Linnik's theorem there are constants $c,L$ such that for every $d \geq 2$ and $1\leq a<d$ with $(d,a) = 1$ the least prime $p_{\text{min}}(d,a)$ congruent $a$ modulo $d$ satisfies $$ p_{\text{min}}(d,a) \leq c d^L. $$ Currently the best known value for the exponent is $L=5$ (Xyloris). On the extended Riemann hypothesis or the generalized Riemann hypothesis, we have $L = 2+\epsilon$ for every $\epsilon > 0$. A folklore conjecture (sometimes attributed to Chowla, sometimes to Heath-Brown) states that $L = 1+\epsilon$ for all $\epsilon$.

Fix a prime number $p$ and fix $a$, say $a=1$. For my purposes the only relevant case is when $d$ a power of the fixed prime number $p$. In this case stronger results are known. Let $L(p)$ be defined as $$ L(p) = \limsup_{j \to \infty} \frac{\log(p_{\text{min}}(p^j,1))}{j \log(p)}. $$ In other words $L(p)$ is the infimum over all real numbers $L> 0$ such that $p_{\text{min}}(p^j) \leq c_L p^{jL}$ for some $c_L > 0$ and all $j \geq 1$. Barban, Linnik and Tshudakov proved $L(p) \leq \frac{8}{3}$. Gallagher established $L(p) < 2.5$ and Huxley improved this to $L(p) \leq 2.4$. The best bound I am aware of can be found in a paper of Banks-Shparlinski: $L(p) < 2.1115$.

My question is: what can be said if $\limsup$ is replaced by $\liminf$? Let's define $$ K(p) = \liminf_{j \to \infty} \frac{\log(p_{\text{min}}(p^j,1))}{j \log(p)}.$$ Clearly, $K(p) \leq L(p)$. So according to the strongest conjectures on $L(p)$ one would have $K(p) =1$. However, it seems possible that one can approach $K(p)$ with different methods. Put differently: one "only" needs to show that for infinitely many $j$ there is a small prime in the arithmetic progression $\equiv 1 \bmod p^j$

**Question:** Are there upper bounds for $K(p)$ which are better than the known bounds for $L(p)$?

(For me the case $a = 1$ is sufficient, but I don't see how this might be useful.)