Phenethylhydrazone decomposition mechanism

I was going through a problem in the book Advanced problems in organic chemistry by Chouhan:

I figured out that since di-isopropyl amide will act as a hindered base it will take out the following hydrogens (in the form of $ce{H+}):$

1. the one where the carbanion formed is in conjugation with benzene ring
2. the one where carbanion formed is in conjugation with nitrogen.

After that these carbanion(s) will attack respectively on their adjacent $ce{C-C}$ σ-bonds and the nitrogen(s) will depart in the form of $ce{N2},$ this gives me option (a) as the answer, but the solution manual proposes a different reaction mechanism:

I am not able to understand the formation of the three-membered ring, please help.
Is there a mistake in my thinking or are there some other concepts involved?

I am an undergraduate student of organic chemistry.