This question is built on drawn structures, so the options cannot be reproduced as text. The chemistry it tests is set out below.

A quaternary trimethylammonium group is the signature of Hofmann elimination. The -N⁺Me₃ unit is an excellent leaving group as the neutral amine NMe₃, and on treatment with base the substrate undergoes E2 elimination.

What makes Hofmann elimination distinctive is its regiochemistry. Ordinary E2 reactions with small leaving groups follow Zaitsev's rule and give the more substituted, more stable alkene. Here the opposite happens: because the ammonium leaving group is bulky and positively charged, the base is directed to the least hindered β-hydrogen, and the product is the less substituted (terminal) alkene — the Hofmann product.

Two further features are worth noting. The transition state requires an anti-periplanar arrangement of the β-hydrogen and the leaving group, so conformational access decides which hydrogen can be removed in a cyclic substrate. And the reaction is a genuine one-step E2, so no carbocation forms and no skeletal rearrangement occurs.

The correct option is therefore the alkene formed by removing a proton from the less substituted β carbon, with trimethylamine expelled; options showing the Zaitsev (more substituted) alkene, or a substitution product from attack at carbon, do not reflect this chemistry.

Per the official final answer key the answer is option (C).