 Solution

A coherent mechanism is an allene cyclization–norbornene insertion–cyclopropane closure cascade. Label the propargyl chain $\mathrm{N}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}\equiv \mathrm{CH}$ as $D-E-F$. Base-assisted propargyl-to-allenyl isomerization gives $\mathrm{N}-\mathrm{CH}=\mathrm{C}=\mathrm{C}{\mathrm{H}}_2$. The hydrogen relocation is important: the terminal carbon must become a methylene before the final C–H substitution.

Reduction of the palladium precursor gives a phosphine-ligated Pd(0) catalyst. Oxidative addition into the aryl C–I bond supplies an aryl–Pd(II) intermediate. Intramolecular carbopalladation of the allene forms the aryl–$E$ bond and the six-membered lactam ring, retaining $D=E$ and placing Pd on the terminal carbon $F$. This allylic organopalladium connectivity may be represented by its interconverting sigma and pi forms.

Norbornene approaches from its less hindered exo face and inserts into the $F$–Pd bond. Carbon $F$ bonds to one former norbornene alkene carbon, while Pd bonds to the adjacent one. The resulting intermediate is geometrically unable to undergo an ordinary favorable beta-hydride elimination: the relevant syn arrangement is blocked, or would produce a bridgehead alkene.

Base-assisted activation of an $F$–H bond gives a cyclopropane-forming dialkylpalladium connectivity, represented as a four-membered Pd–C–C–$F$ ring. Reductive elimination closes the bond between $F$ and the Pd-bearing norbornene carbon, releases Pd(0), and gives the cyclopropane. The lactam alkene $D=E$ remains intact. Carbonate removes the acid equivalent generated in the overall arylation.

This mapping accounts for the six-membered heterocycle, the new three-membered ring and exo addition to norbornene. The catalyst and substrates are achiral, so the drawn relative stereochemistry represents a racemic product. This is a proposed organopalladium pathway; the question's product alone does not uniquely establish the lifetime or order of every palladium intermediate.