I have to prove that every projective module has a free complement.
Now Rotman ask it to first do for $R=Z/6Z$ and $P=Z/2Z$, We know $Z/6Z \cong Z/2Z \oplus Z/3Z$. Now $Z/2Z$ is projective as it a summand of a free module, namely $_RR$ so choosing free complement as $F=Z/6Z$ will work here such that $Z/6Z \cong Z/2Z \oplus F \cong Z/2Z \oplus Z/6Z$. Is it correct?
How to go for general proof?
If $P$ is projective and $Q$ is any complement of $P$ in a free module,then $P\oplus Q\oplus P\oplus Q\oplus P\oplus Q\oplus\cdots$ is a free complement.
In your example, no finitely generated free module is a complement to $P$: indeed, every f.g. free module has $6^n$ elements for some $n$ and $P$ has $2$, so that the direct sum of $P$ and a free module has $2\times 6^n$ elements, and it is therefore never free. In particular, your argument is not correct.