 Nonregular language satisfying the pumping lemma

Let $K\subseteq \mathbb{N}$ be undecidable andPumping its first symbol always stays in $L$, so it satisfies the pumping lemma with pumping length one. Yet $L\cap a{b}^{\ast }=\{a{b}^n:n\in K\}$ is not regular, because a finite automaton for it would decide $K$.