= Solution
Let $\mathrm{Ar}=\mathrm{4{-}NO_2C_6H_4}$ and number the carbon substituent from <oxygen>: $\mathrm{ArO{-}CH_2(1){-}CH_2(2){-}CH_2(3){-}CN}$. <sodium hydride> removes a <proton> from C3, alpha to the nitrile, forming <hydrogen> and a nitrile-stabilized <carbon> <anion>. That <carbon> attacks the nitroaryl ipso <carbon> bearing O. The tether closes a five-membered cyclic Meisenheimer intermediate with sequence \b[C3–C2–C1–O–aryl ipso <carbon>]. The para <nitro group> stabilizes the negative charge generated by temporary loss of <aromaticity>.
Rearomatization breaks the aryl–O bond, releasing the tether <oxygen> as an <alkoxide> while retaining the new aryl–C3 bond. <protonation> gives \b[$\mathrm{Ar{-}CH(CN){-}CH_2{-}CH_2OH}$]. This is a Truce–<Smiles rearrangement>: the aryl group migrates from <oxygen> to <carbon>. The nitrile remains attached to its original side-chain <carbon>; the original <ether> <oxygen> becomes the terminal <alcohol> <oxygen>. The substrate and product have identical atomic composition. Ordinary substitution at a saturated chain <carbon> would not produce this aryl–<carbon> bond or restore the terminal OH.
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