A non-regular language satisfying the pumping lemma
You are not supplying the definition of $L$, so here is one definition which works:

$$ L = \{ a b^n c^n : n \geq 0 \} \cup \{ a^k w : k \neq 1 \text{ and } w \in \Sigma^* \text{ doesn't start with $a$} \}. $$

Choose a pumping length of $2$. A word $ab^nc^n \in L$ can be pumped by pumping the $a$ part. A word $aaw$ can be pumped by pumping the $aa$ part. Finally, a word $bw$ or $cw$ can be pumped by pumping the $b$ part or $c$ part.

Your definition seems to be

$$ L' = \{ a b^n c^n : n \geq 0 \} \cup \{ a^k w : k \neq 2 \text{ and } w \in \Sigma^* \}, $$

but for this language the pumping lemma does work. Indeed, apply it to $ab^nc^n \in L'$ for large enough $n$. Then $ab^nc^n = xyz$ so that $|y| \geq 1$ and $xy^iz \in L'$ for all $i \geq 0$. Considering the case $i = 2$, we see that $y = a$ necessarily. But then, taking $i = 0$ we get $xz = b^nc^n \notin L'$.

Perhaps the exercise meant you to use the version of the pumping lemma in which $i \geq 1$ instead of $i \geq 0$.

We can show that $L$ is not regular in many ways:


Use one of the many general versions of the pumping lemma which can force the $b^nc^n$ to be pumped. For example, the general version on Wikipedia states that there exists $p$ such that any word $uwv \in L$ with $|w| \geq p$ can be partitioned as $w = xyz$ so that $|xy| \leq p$, $|y| \geq 1$ and $uxy^iz \in L$ for all $i \geq 0$. Choose $u = a$, $w = b^nc^n$ for $n \geq p$ and $v = \epsilon$, and follow the usual argument.
If $L$ were regular then $L \cap ab\Sigma^* = \{ ab^nc^n : n \geq 1 \}$ would also be regular, and you can prove that it isn't using the usual pumping lemma.
Use the Myhill–Nerode criterion. The words $ab^n$ are pairwise inequivalent, and so $L$ is not regular.