$P$ is projective . Show that $P$ is a direct summand of a free $R-$ module $P$ is projective . Show that $P$ is a direct summand of a  free $R-$ module
I am using the definition of a projective module as $P$ is projective if every exact sequence $M\rightarrow P\rightarrow 0$ splits 
How to proceed?
 A: Choose generators of $P$ to get a surjection from a free module $F$ onto $P$. This gives you a split (by your definition!) exact sequence $0 \to K \to F \to P \to 0$, showing $F = K \oplus P$.
A: Since people are rejecting my edits suggesting to write an answer, I do it here:
Let me first reformulate the question to contain all relevant details:

Assume as definition of projective module the following:
$P$ is projective if every exact sequence $M\rightarrow P\rightarrow
 0$ splits, that is, for every epimorphishm $f:M \to P$ there exists a
  right inverse homomorphism $f':P \to M$ such that $ff' = 1_P$
Then, I need to prove the following implication:
If $P$ is projective, show that $P$ is a direct summand of a free
  $R$-module, that is, there exists a free module $F$ such that $F = P
 \oplus T$ for some other module $T$.

Then let me add all the details that were neccessary to me based on @MooS' answer:
Choose generators of $P$ to get a surjection from a free module $F$ onto $P$. More precisely, prove that:

If a module $M$ has a spanning set $X \subseteq M$ then there is an
  epimorphism $R^{(X)} \to M \to 0$ 

If you need it, you can find the proof in the book Rings and Categories of Modules, theorem 8.1. 
Then what you have is a exact sequence $F \stackrel{f}\to P \to 0$ and taking the kernel $K$ of $f$ we get an exact sequence $0 \to K \stackrel{i}\to F \stackrel{f}\to P \to 0$.
Since $P$ is projective, by your definition, there exists $f'$ such that $ff' = 1_P$ but this is one characterization of being a splitting sequence. Another characterization of splitting sequence is that the center is sum of the non-null extremes, showing that $F = K \oplus P$.
