Irregularity of $\{a^ib^jc^k \mid \text{if } i=1 \text{ then } j=k \}$
The pumping lemma can only be used to show that a language is not regular, not that it is regular. This is, I think, the fundamental misunderstanding here... that you are unable to use the pumping lemma to prove a language irregular does not constitute a proof of irregularity.

A language is regular if, equivalently,


it can be accepted by an NFA
it can be generated by a regular grammar
it can be described by a regular expression
it can be partitioned into finitely many equivalence classes under the indistinguishability relationship (as per the Myhill-Nerode theorem)


This last point can usually be pretty helpful in cases where the pumping lemma fails. Here, you merely need to demonstrate that there are infinitely many equivalence classes for strings in your language; you could argue that strings of the form $ab^nc$ constitute separate classes for all $n$.

Furthermore, as pointed out in the comments, you can often use closure properties of language classes in order to demonstrate that a particular language isn't in the class. For instance, you know that regular languages are closed under intersection. Using that fact, you could show that your language can't be regular, since if it were, the language $ab^nc^n$ would be regular (a contradiction, since you know that language isn't regular... you could even prove this easier version with the pumping lemma).

EDIT: Just in case how to apply the pumping lemma here is unclear, I'll give an example.

Choose the string $uvw = ab^pc^p$. Since $|uv| &lt; p$ and $v &gt; 0$, $v$ must be some non-empty string possibly beginning with a single $a$, followed by multiple $b$'s. If $v$ contains the $a$, pumping will cause there to be a different number of $a$'s, so we get a string not in the language. If the string does not contain the $a$, it consists only of $b$'s, and pumping will cause the number of $b$'s to be different from $n$. The resulting string is not in the language since the number of $b$'s is not equal to the number of $c$'s. Therefore, this string cannot be pumped to another string in the language, a contradiction of the assumption the language was regular; hence, the language is not regular.