Piperidine is exhaustively methylated ($\mathrm{CH}{}_3\mathrm{I}$ in excess, then $\mathrm{Ag}{}_2\mathrm{O}$/$\mathrm{H}{}_2\mathrm{O}$, heat) and the resulting tertiary amine is subjected to the whole sequence once more. The hydrocarbon finally obtained, together with trimethylamine, is:

The first round gives N,N-dimethylpent-4-en-1-amine by ring-opening Hofmann elimination (the ring N-C bond breaks, the C=C forms at the terminal position). The second round forms the quaternary ammonium salt, and elimination again removes the most accessible beta-hydrogen, giving $\mathrm{CH}{}_2=\mathrm{CH}-\mathrm{CH}{}_2-\mathrm{CH}=\mathrm{CH}{}_2$ (penta-1,4-diene) and $\mathrm{N}(\mathrm{CH}{}_3){}_3$.