Proving that a language satisfies the pumping lemma condition
The language is regular since it is the language of the regular expression $(aa)^*(bb)^*$.

Every regular language satisfies the conditions of the pumping lemma – this is precisely the statement of the pumping lemma! So one way to show, from first principles, that this language satisfies the conditions of the pumping lemma would be to just repeat the proof of the pumping lemma.

However, in this case we can show this more directly. We will show that the pumping lemma holds with a pumping constant of $p = 2$. Suppose that $w = a^{2i} b^{2j} \in L$. Note that $|w| \geq 2$. We consider two cases:


Case 1: $i \neq 0$. In this case, we can break $w$ in the following form: $x = \epsilon$, $y = a^2$, $z = a^{2(i-1)} b^{2j}$. We have $w = xyz$, $|xy| = 2 \leq p$ and $y \neq \epsilon$. For every $n$, $xy^nz = a^{2(i-1+n)} b^{2j} \in L$.
Case 2: $i = 0$. In this case, $j \neq 0$. We break $w$ into $x = \epsilon$, $y = b^2$, $z = b^{2(j-1)}$. The rest of the proof is very similar.