= Solution
The key sequence is \b[deprotonation, <anionic eight-pi electrocyclization>, then elimination]. Label the conjugated route from the lower cyclopentene to the right cyclopentene as
$$
C_1=C_2-C_3-C_4=C_5-C_6=C_7,
$$
where $C_3$ is the methylene joining the lower ring to the central alkene, and $C_4$ carries the $\mathrm{CH_2OC(O)NMe_2}$ substituent.
<potassium hexamethyldisilazide> removes a <proton> at $C_3$. The resulting anion is delocalized over the seven-carbon array: its six pi-bond <electrons> plus the lone pair make an eight-electron system. Thermal <electrocyclic reaction> is disrotatory for this $4n$ <electron> count. Joining $C_1$ and $C_7$ forms the seven-membered central ring and puts the remaining anionic character at $C_4$, with double bonds at $C_2=C_3$ and $C_5=C_6$.
That $C_4$ anion then forms $C_4=CH_2$, expelling the O-dimethylcarbamate group from the attached methylene. This <E1cb elimination> supplies the exocyclic methylene and explains why no oxygen remains in the depicted product. The disrotatory closure, with the cyclopentene ends folded to avoid unfavorable overlap, gives the trans relationship between the two lower fused-ring bonds shown by the wedge and dash. A simple \[3,3\] rearrangement would not create the required bond between the two remote cyclopentene termini.
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