The compound shown is one stereoisomer of bicyclo[2.2.1]heptan-2-ol (norbornan-2-ol), drawn with wedges. Number the bridgehead next to the $\mathrm{CHOH}$ carbon as C1, the $\mathrm{CHOH}$ carbon as C2 and the other bridgehead as C4. Its configuration is

C2: priorities $\mathrm{OH}>\mathrm{C}{}_1$ (attached to C,C,C) $>\mathrm{C}{}_3$ (attached to C,H,H) $>\mathrm{H}$; the drawn arrangement gives $(S)$. C1: its three ring neighbours are C2 (O,C,H), C6 (C,H,H) and C7 (C,H,H); C2 is first, and C6 versus C7 is decided at the next sphere (C6 leads to C5 which bears C4, while C7 leads directly to C4, whose set (C,C,H) beats the (C,H,H) of C5), so the order is $\mathrm{C}{}_2>\mathrm{C}{}_7>\mathrm{C}{}_6>\mathrm{H}$, giving $(S)$. C4 with $\mathrm{C}{}_3>\mathrm{C}{}_7>\mathrm{C}{}_5>\mathrm{H}$ gives $(R)$. The compound shown is $(1S,2S,4R)$; option B is its enantiomer.