 Solution

Label the starting exomethylene carbon $a$, the aryl-bearing carbon $b$, the ethyl CH${}_2$/CD${}_2$ carbon $c$, and the terminal methyl $d$:Insertion into Ni-H places H on $a$ and Ni on $b$. Elimination of a beta H or D from $c$ gives $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}(\mathrm{Ar})=\mathrm{C}(\mathrm{L})-\mathrm{C}{\mathrm{H}}_3$ and Ni-H or Ni-D. Thus the labeled substrate can retain one D on the new vinylic carbon while exporting the other into the catalyst hydride pool.

Ni-D can then add to the exomethylene of either aryl substrate, introducing D into the methyl group derived from $a$. Reversible insertion/elimination and substrate exchange also introduce D into the vinylic position of the originally unlabeled compound. This explains both the approximately comparable methyl labeling of the two products and the larger residual vinylic labeling of the initially CD${}_2$-bearing substrate.

The decisive observation is intermolecular crossover of deuterium into the unsubstituted-phenyl product. It rules out a purely intramolecular, identity-preserving 1,3-H/D shift. The unequal percentages retain kinetic and substrate-history information; they are not a requirement for completely statistical H/D equilibration and cannot determine all elementary rate constants by themselves.