(S)-2-Bromooctane is treated successively with (i) aqueous NaOH (pure SN2), (ii) PBr3 (SN2 at carbon), (iii) NaCN (pure SN2).

Each $S_N2$ step inverts the configuration at C-2. In steps (i) and (ii) the priority order stays the same ($\mathrm{Br}$ or $\mathrm{OH}$ $>$ $\mathrm{C_6H_{13}}$ chain $>$ $\mathrm{CH_3}>\mathrm{H}$), so (S)-bromide $\to$ (R)-2-octanol $\to$ (S)-2-bromooctane (two inversions, net retention). In step (iii) $\mathrm{CN}$ (carbon with N,N,N) outranks the chain carbon (C,H,H), so the order is $\mathrm{CN}>\mathrm{chain}>\mathrm{CH_3}>\mathrm{H}$, corresponding to the same order as with $\mathrm{Br}$. Third inversion converts (S) to (R): (R)-2-methyloctanenitrile.