= Solution
A coherent mechanism is an \b[allene cyclization–<norbornene> insertion–cyclopropane closure cascade]. Label the propargyl chain $\mathrm{N-CH_2-C\equiv CH}$ as $D-E-F$. Base-assisted propargyl-to-allenyl isomerization gives $\mathrm{N-CH=C=CH_2}$. 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 \b[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.
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