As Igor points out, the cross-ratio of a sub*set* of four distinct points is not well-defined. If you meant to stipulate that all *tuples* of four distinct points from $P$ should have different cross-ratios, except for the unavoidable repetitions, then the answer is yes, you can conclude that $f$ is the identity transformation:

Since $f$ is a projective linear transformation, it preserves cross-ratios. In particular, the cross-ratio of $(f(p_1), f(p_2), f(p_3), f(p_4))$ is the same as that of $(p_1, p_2, p_3, p_4)$. Thus, by the hypothesis, $f$ in fact must permute already the set consisting of the first four points, in a manner which leaves the cross-ratio unchanged. If $f$ isn't the identity, then it must permute them in two pairs. Without loss of generality (renumbering points if necessary and conjugating everything with a suitable projective linear transformation to pin down the first three points) the only possibility is swapping $p_1=\infty$ with $p_2=0$ and $p_3=1$ with some $p_4$; thus $f\colon x\mapsto p_4/x$ and (**edit** if $f$ moves $p_5$) $p_6=p_4/p_5$; but then the cross-ratio of $(\infty,0,1,p_5)$ would equal that of its $f$-image $(0,\infty,p_4,p_6)$, a tuple supported on a different subset of $P$.

**Edit** - Or else, this $f$ fixes both $p_5$ and $p_6$, which must then be two distinct square roots of $p_4$ in the field we're working with. But then the cross-ratio of $(\infty,0,1,p_5)$ would equal that of its $f$-image $(0,\infty,p_4,p_5)$; again violating the hypothesis.