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

The two steps together are a hydroboration / Suzuki-Miyaura coupling sequence, one of the standard ways of joining an alkene to an aryl group.

Step (i), BH₃ in THF — hydroboration. Borane adds across the double bond in a single concerted step, so the addition is syn: hydrogen and boron are delivered to the same face. The regiochemistry is anti-Markovnikov, boron going to the less substituted carbon. Two reasons combine to give this: boron is the electropositive, larger partner and prefers the less hindered position, and the four-centre transition state places the developing partial positive charge on the more substituted carbon. The product is an alkylborane.

Step (ii), PhBr with Pd(PPh₃)₄ and Na₂CO₃ — Suzuki coupling. The palladium(0) catalyst undergoes oxidative addition into the C-Br bond of bromobenzene. The carbonate base then activates the alkylborane towards transmetalation, transferring the alkyl group from boron to palladium — the base is essential here, since neutral organoboranes are too weakly nucleophilic to transmetalate. Reductive elimination finally forms the new carbon-carbon bond and regenerates Pd(0).

The overall outcome is that a phenyl group becomes attached at the less substituted carbon of the original alkene, with the methoxy substituent carried through unchanged. Both steps proceed with retention of configuration at carbon, so the syn stereochemistry set in the first step is preserved.

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