The major product formed in the following reaction is:
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\(\mathrm{SmI_2}\) reduces the ketone to a ketyl radical that cyclizes onto the tethered \(\alpha,\beta\)-unsaturated ester through carbon, not oxygen; count the chain length to see it is a 5-exo-trig cyclization, giving a fused bicyclic alcohol with a pendant \(\mathrm{CH_2CO_2Me}\) group after the \(\mathrm{MeOH}\) quench.
$\mathrm{SmI_2}$ is a one-electron reductant, and the classic move it makes with a ketone is to hand over an electron and generate a ketyl radical. When that ketone has an alkene hanging off it on a chain of the right length, the ketyl radical reaches over and forms a new ring by attacking the alkene. That is exactly the setup in this question.
Where the radical starts: the ketone carbon of the cyclopentanone becomes a radical, ketyl, after picking up one electron from $\mathrm{SmI_2}$.
Where the radical goes: tracing the chain from that carbon, through the ring's substituted carbon, then two $\mathrm{CH_2}$ groups, to the nearer carbon of the $\mathrm{C{=}C{-}CO_2Me}$ alkene, that is a five-atom path. Radical cyclizations strongly prefer forming a five-membered ring this way, 5-exo-trig, over a six-membered one, so the ketyl radical attacks the nearer alkene carbon, not the one next to the ester.
What is left after the new ring closes: closing that five-membered ring uses up the double bond, leaving the radical, and eventually after a second reduction an anion stabilized by the ester, sitting on the carbon that was attached to $\mathrm{CO_2Me}$. That carbon becomes a simple $\mathrm{CH_2}$ once it picks up a proton.
Where the proton comes from: $\mathrm{MeOH}$, added after the $\mathrm{SmI_2}$ step, supplies the proton both to that ester-stabilized carbon, giving $\mathrm{{-}CH_2CO_2Me}$, and indirectly through the samarium alkoxide, to the oxygen that was originally the ketone, giving a tertiary alcohol at the ring-fusion carbon.
Put together, the two original rings, the cyclopentanone ring and the newly closed five-membered ring, are now fused at the carbon that used to be the carbonyl carbon, which carries the new $\mathrm{OH}$. The pendant group on the new ring is $\mathrm{{-}CH_2CO_2Me}$, not a ring oxygen and not an ester carbon fused directly into the ring. That rules out the two oxygen-in-ring answers, they would need the oxygen, not the carbon, to attack the alkene, and the direct-fused-ester answer, it would need a 6-exo cyclization, which loses out to the favored 5-exo pathway here.
Let's summarize:
$\mathrm{SmI_2}$ converts the ketone to a ketyl radical, which cyclizes onto the tethered alkene, at carbon, not oxygen.
The tether length favors a five-membered ring, 5-exo-trig, giving a fused bicyclic, hydrindane-type, skeleton.
$\mathrm{MeOH}$ quench delivers protons to give a tertiary alcohol at the ring fusion and a $\mathrm{{-}CH_2CO_2Me}$ pendant group.
The correct product is the fused bicyclic alcohol with the pendant ester arm, option (B).
