Pumping Lemma for $\mathcal{L} = \{ \omega \omega^R a^{|\omega|} : \omega \in \{a,b\}^* \} $
The problem is not very hard if you break the proof of the pumped string not being in the language into cases. So let's work this out.
$$L = \{\omega \omega^Ra^{|\omega|}\;| \; \omega\in \{a,b\}^*\}$$
For every $n$, let us consider the string $\omega = a^nb^nb^na^na^{2n}$. You can check that this does belong to the language.
We are supposed to show that for any partition $\omega=uvwxy$, such that
$$|vwx|\leq n \;\mbox{and} \; |vx|\geq 1$$ there exists $k$ such that $uv^kwx^ky \notin L$.
Case1: $v,x$ lie inside the first $a^n$. Since, $|vx|\geq 1$, we can see that for $k=0$, the resulting string will not be in $L$.
Case2: $v,x$ lie in the starting $a^nb^n$. You should notice that if you again pump the substrings with $k=0$, then the substring cannot be broken into $\omega\omega^R a^{|\omega|}$, because the count of $a$'s at the end remains the same, and we cannot have an $\omega$ of length $2n$ in the resultant string for it to happen.
Case3: $v,x$ lie inside the first $b^n$. Here, again, you can see how $k=0$ will pump the string outside the language.
The further cases are similar to what described above. Work it out for better understanding. Please let me know if you aren't comfortable with any part of this solution.
PS: The $a^{|\omega|}$ which makes you sick was the sole reason this language is not a CFL, XD.