= Solution
Treat this as standalone <organophosphorus chemistry>. The phosphorus(III) lone pair attacks the terminal nitrogen of the <organic azide>. Cyclization of the initial phosphazide and loss of nitrogen give an <iminophosphorane>:
$$
\mathrm{ArN_3+PPh_2(C_6H_4CONH_2)
\longrightarrow ArN=PPh_2(C_6H_4CONH_2)+N_2}.
$$
The ortho <amide> provides a neighboring acidic NH <proton> and a <carbonyl> oxygen donor. A <proton> relay to the imino nitrogen and intramolecular O$\to$P attack generate a five-membered, O-bound pentacoordinate phosphorus intermediate. Water can then attack phosphorus, with <proton transfer> to nitrogen, followed by P–N cleavage and opening of the neighboring-group contact.
The overall hydrolysis products are
$$
\boxed{\mathrm{ArNH_2+Ph_2P(O)(C_6H_4CONH_2)+N_2}.}
$$
The <carbonyl> group assists the hydrolysis and is regenerated; the product is \b[an aniline and the ortho-carboxamide <phosphine oxide>], rather than an amide-linked ligation product. The favorable five-membered intramolecular interaction and <proton> delivery help hydrolyze a normally rather persistent aryl <iminophosphorane>. This explanation concerns only the printed chemical transformation.
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