= Solution
Phosphine binding breaks the hydroxo-bridged dimer into mononuclear Rh-OH units. <Hydroxide>/boronic-acid association forms a Rh-O-B linkage; transfer of the aryl group from boron to rhodium, with Rh-O bond cleavage, gives the Rh-aryl initiator. This is <transmetalation>, with rhodium remaining formally Rh(I).
Writing $\mathrm{Ar}=p\!-\!\mathrm{C_6H_4Me}$ and $L=\mathrm{PPh_3}$, the balanced net activation is
$$
\mathrm{[(COD)Rh(\mu-OH)]_2+2ArB(OH)_2+2L
\longrightarrow2(COD)Rh(L)Ar+2B(OH)_3}.
$$
The stoichiometric byproduct is \b[boric acid], $\mathrm{B(OH)_3}$. Each monomeric Rh(I) product is a 16-electron complex: $d^8$ Rh(I), four <electrons> from COD, two from phosphine and two from the aryl X ligand. Temporary phosphine dissociation makes room for alkyne binding during propagation.
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