In this theorem finally I'm able to prove that the sequence $0\xrightarrow{} K_1\xrightarrow{\phi}K_2 \oplus P_1 \xrightarrow{\psi}P_2\xrightarrow{} 0 $ is exact. Then it is written that clearly as $P_2$ is projective so proof is done. But I'm unable to understand how it is clear ? How $P_2$ projective implies $K_1\oplus P_2 \cong K_2\oplus P_1$ ?
Can anyone explain it please ?
If $P_2$ is projective, then the short exact sequence you've written down is (right) split, and hence the middle term is isomorphic to the direct sum of the outer terms.
Indeed, to say any short exact sequence whose last non-zero term is $P_2$ is split is equivalent to saying $P_2$ is projective. See here.