 Solution

Let $K\subseteq \mathbb{N}$ be the undecidable set denoted $\mathbb{K}$ in the question. The language satisfies the pumping lemma for regular languages with pumping length one. For any nonempty $s\in L$, write $s=xyz$ with $x=\epsilon$, $y$ its first symbol, and $z$ the remaining suffix.

If $s\in b^{\ast }$, then $x{y}^{i}z\in b^{\ast }$ for every $i\ge 0$. Otherwise $s=wa{b}^n$ for some $n\in K$. If the chosen first symbol lies in $w$, pumping it preserves the same final suffix $a{b}^n$. If $w=\epsilon$, then $s=a{b}^n$: pumping down gives $b^n\in b^{\ast }$, while every positive pump gives $a^i{b}^n$ and all but its last $a$ can be absorbed into $w$. Thus every sufficiently long word has a valid pumping decomposition.

Nevertheless, if $L$ were regular language, closure under intersection would makeregular. A finite automaton deciding this unary language would decide whether $n\in K$ by running it on $a{b}^n$, contradicting the halting problem. ThereforeThis is a nonregular language satisfying the pumping lemma.