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The Needham-Schroeder protocol is:
1 A→S:A,B,Na
2 S→A:{Na,B,Kab,{Kab,A}Kbs}Kas
3 A→B:{Kab,A}Kbs
4 B→A:{Nb}Kab
5 A→B:{Nb−1}Kab

Here Na and Nb are nonces.
Kab is the key between a and b
Kas is the key between A and KDC
Kbs is the key between B and KDC

My Question: In step 2 if we don't encrypt with Kbs & in step 3 we encrypt
using Kab instead of Kbs what will happen? Following are the changes:
2 S→A:{Na,B,Kab,{Kab,A}}Kas
3 A→B:{Kab,A}Kab

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up vote 2 down vote accepted

Kab is session key, which generated by KDC and must be new every time. So the most simple answer is: Bob could not to decrypt {Kab,A}Kab, because he does not know Kab before step 3.

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