Why is $L := \{b^2a^nb^ma^3|m,n \geq 0\}$ a regular language?
The pumping lemma says that there is some&nbsp;$p$ such that every string longer than&nbsp;$p$ can be split in some way that can be pumped.

You've shown that there's some&nbsp;$p$ and some splitting that doesn't work but (since the language is regular), that just means you chose the wrong&nbsp;$p$ and/or the wrong splitting of your string.

In fact, the value of&nbsp;$p$ is the number of states in the smallest automaton that accepts the language (plus one, I think). The language you give is going to need about eight or nine states, so your&nbsp;$p$ was too small. However, that doesn't much matter. Given the string $b^2a^{p-3}ba^3$, we can split it as $x=b^2$, $y=a$, $z= [\text{the rest}]$ and now $xy^iz\in L$ for all $i\geq 0$, as the pumping lemma claims.  Sure, some other splits don't work, but that one does and the lemma just claims that some split works.

Remember: to show that a language is non-regular using the pumping lemma, you need to show that, for every&nbsp;$p$, there is at least one word of length at least&nbsp;$p$ in the language such that every valid splitting as $xyz$ results in $xy^iz\notin L$ for at least one&nbsp;$i$.