How to prove $\{a^nb^na^n \mid n\geq1\}$ is not regular using pumping Lemma Here the problem is that I’m confused how to take the pumping value $p$ is it arbitrary any value?
Also I don’t know if I should prove all $3$ conditions of the pumping lemma is false or if any one should be proved as false by considering the given language as regular by proof by contradiction.
Let's recall the Pumping Lemma statement:

For all regular language $ L $, exists a positive constant called $ p $ (i.e. $ p \geq 1 $) such that every string $ w $ in $ L $ with length $ |w| \geq p $ exist strings $ x, y, z $ such that $ w = xyz $ and that statisfying the following conditions:

$ |y| \geq 1 $
$ |xy| \leq p $
$ xy^iz \in L, \: \forall i \geq 0 $


So yes, $ p $ is an arbitrary value and a fixed costant, called &quot;pumping length&quot;. The proof for your language may be the following:
Let's assume that $ L $ is a regular language, so for all strings $ w \in L $ it must satisfy conditions of PL.
Now choose a string $ w $ that is in $ L $, let choose $ w = a^pb^pa^p $ and we can split $ w $ in $ xyz $, so let's say that:

$ x = a^k $
$ y = a^h $
$ z = a^{p-h-k}b^pa^p $

with $ h \geq 1 $ and $ h + k \leq p $.
Now note that the first two conditions of the PL are satisfied, but for the third, if we take $ i = 0 $ then $ w = xy^0z = a^ka^{p-h-k}b^pa^p = a^{p-h}b^pa^p $ and this string clearly doesn't belongs to $ L $ and this contraddict the initial assumption of $ L $ regular, so this meanings that $ L $ isn't a regular language because it doesn't satisfy the PL.