ECFP harder than ECDLP?
It's not harder.


Solve the DL problem to get $x$ in $Q=x.P$.
$Q = \sum_{i=1}^{n} x_i.P = (\sum_{i=1}^{n} x_i).P= x.P$


=> any set of $x_i$ that sums to $x$ (modulo the group order) is a solution to ECFP.