I understand step A involves reductive ozonolysis. and that step B is addition involves addition of ammonia to the carbonyl site to form imines. But what does the sodium salt of trichloroacetic acid do? I'm inclined to say C or D since I cannot explain how else the extra double bonds form with a strong slant towards C because I would expect di-chlorination for D.
Step 1 is ozonolysis with reductive work-up giving butan-1,4-dialdehyde. Step 2 reaction with NH3 gives pyrrole. Step 3 is the interesting one. As the discussion in the comments states the thermal decomposition of sodium trichloroacetate gives dichlorocarbene. This adds to one of the double bonds of pyrrole to give initially the dichlorocyclopropane. This is unstable and rearranges to give 3-Chloropyridine. This gives higher yields in neutral aprotic solvents reference here. Further discussion and mechanism here