is this language regular and why pumping lemma doesn't work?
It's a "trick" question. The language is regular because
\begin{align*}
\{aba^{\mathrm{R}}\mid a,b\in\{0,1\}^*\}
    &amp;= \big\{\varepsilon b\varepsilon^{\mathrm{R}}\mid b\in\{0,1\}^*\big\}
       \cup \big\{a b a^{\mathrm{R}}\mid a\in\{0,1\}^+,\ b\in\{0,1\}^*\big\}\\
    &amp;= \{0,1\}^*
       \cup \big\{a b a^{\mathrm{R}}\mid a\in\{0,1\}^+,\ b\in\{0,1\}^*\big\}\\
    &amp;= \{0,1\}^*\,.
\end{align*}