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Lindenbaum theorem states that any consistent set of sentences can be enlarged to a maximal consistent set of sentences. Thus, by choosing a consistent set of sentences in Peano arithmetic and by Lindenbaum theorem, they can be enlarged to a complete theory. However, this contradicts that Peano arithmetic is incomplete. Any explanation?

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There's no contradiction at all: the completion is not the same as the theory being completed.

Peano arithmetic (by which I presume you mean first-order Peano arithmetic) $\mathsf{PA}$ is consistent and incomplete. Lindenbaum says that since $\mathsf{PA}$ is consistent, there is a consistent complete $T$ with $\mathsf{PA}\subseteq T$. This $T$ is not the same as $\mathsf{PA}$ itself, so there's no contradiction between $T$ being complete and $\mathsf{PA}$ being incomplete. (Also note that $\mathsf{PA}$ has many different complete consistent extensions.)

Now $\mathsf{PA}$ does have a "weak non-completeability" property: no computably axiomatizable consistent extension of $\mathsf{PA}$ can be complete (this result is Rosser's improvement of Godel's first incompleteness theorem, and this property is called "essential incompleteness"). However, this doesn't contradict Lindenbaum, it merely means that the above $T$s cannot be computably axiomatizable.

Noah Schweber
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    Beat me to it again – HallaSurvivor Apr 06 '21 at 05:53
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    @HallaSurvivor Muaha - I say - ha. :P – Noah Schweber Apr 06 '21 at 05:53
  • @Noah Schweber, I think there is a problem. Peano arithmetic is incomplete because it contains some undecidable predicates. The completion of PA (actually) ignores these undecidable predicates and so there is no $\mathsf{PA}\subset T$. –  Apr 14 '21 at 16:54
  • @hermes "Peano arithmetic is incomplete because it contains some undecidable predicates." What? No. There are sentences which are PA-undecidable, but these don't remain undecidable when we pass to a stronger theory. There is no problem here. – Noah Schweber Apr 14 '21 at 16:56
  • Let's say, how do undecidable sentences in Godel's original paper become decidable in a stronger theory? –  Apr 14 '21 at 16:58
  • @hermes By being decidable in that stronger theory. There's no particular mystery here: there's no reason for the Godel sentence for one theory to be undecidable in a different theory. And in a complete extension of $\mathsf{PA}$ they all become decidable. For example, if we look at the completion $\mathsf{TA}=Th(\mathbb{N})$ we get a theory which proves every consistent r.e.-axiomatizable theory's Godel(-Rosser) sentence; as $\mathsf{TA}$ isn't arithmetically definable, it itself doesn't have a Godel sentence so there's no contradiction. – Noah Schweber Apr 14 '21 at 16:59
  • This is where I suspect. "as TA isn't arithmetically definable, it itself doesn't have a Godel sentence so there's no contradiction. – ", does not provide an exact solution to undecidable predicates but simply bypass it. –  Apr 14 '21 at 17:05
  • @hermes "does not provide an exact solution to undecidable predicates but simply bypass it" I don't understand your issue here. What do you mean by "an exact solution to undecidable predicates?" What's wrong with "simply bypassing" it? I really don't understand what you're getting at here, and I don't think this comment thread is a good place to hash it out; if you still don't understand my answer you should ask a separate question about it. – Noah Schweber Apr 14 '21 at 17:07
  • What I mean is that it does not touch the nature of undecidable predicates. The true nature of them relies on multi-valued logic. I can not explain more here. It has nothing to do with your answer which remains correct. Just the theory of complete extension could be inadequate. –  Apr 14 '21 at 17:09