Initially $P$ has 5 valence electrons.
Work Step by Step
1) The electron configuration of $P$ is $$[Ne]3s^23p^3$$ $P$ has 5 valence electrons. The electron configuration of $F$ is $$1s^22s^22p^5$$ $P$ has 7 valence electrons. 2) Here we see that both $P$ and $F$ need more electrons to fill an octet ($P$ needs 3 more electrons and $F$ needs 1 more electron), so we would use covalent bonding in $PF_3$ formation, which means each atom must share its own electrons so that all can reach an octet. Since each $F$ atom only needs 1 more electron to reach an octet, it will share 1 electron and $P$ also shares 1 electron so that these 2 electrons would be used by both $P$ and $F$ atoms. Therefore, both $P$ and $F$ atoms would acquire 1 more electron to its atom, and now $F$ atom reaches an octet. The same process applies in forming covalent bonding of $P$ with the other 2 $F$ atoms. So eventually, $P$ atom would give out 3 out its 5 valence electrons to the sharing. However, since each sharing has 2 electrons which can be used by both atoms, after 3 sharings $P$ atom now has 6 bond-sharing valence electrons, plus 2 nonbonding electrons and reaches an octet. The diagram of formation of $PF_3$ is shown below. (a) As we see from the electron configuration of $P$, the valence electrons in the $3s$ and $3p$ subshells. There are 2 electrons in subshell $3s$ and 3 electrons in subshell $3p$. Therefore, initially $P$ has 5 valence electrons.