Proving that a certain language is regular using pumping lemma
You cannot use the pumping lemma to show that a language is regular. The pumping lemma gives a property of regular languages: if $L$ is regular then $L$ can be &quot;pumped&quot;. You can use this to show that $L$ is not regular: if $L$ cannot be pumped, then it is not regular. But you cannot use it to show that $L$ is regular: if $L$ can be pumped, it doesn't follow that $L$ is regular; moreover, there are examples of pumpable languages which are not regular. There are extensions of the pumping lemma which can be used to show that a language is regular, but they are not so useful in practice.
In your case, every word in $L$ contain $0$ or $4$ many $a$s, $0$ or $8$ many $b$s, and so on. In particular, it contains at most $4$ many $a$s, at most $8$ many $b$s, and so on. Therefore every word in $L$ has length at most $4+8+\cdots+2^{27}$. Consequently, $L$ is finite. Every finite language is regular.