P vs NP: Assuming P = NP
Yes.  Every* problem in $P$ can be reduced to any other problem in $P$ using polynomial-time reductions (since the reduction is allowed to do polynomial work, it can just solve the problem itself).

If $P=NP$, then it follows that every* problem in $NP$ can be reduced to any* other problem in $NP$ -- in other words, every* problem in $NP$ is $NP$-complete.  This implies that if $L \in P$ and if $P=NP$ then* $L$ is $NP$-complete.

Footnote *: There is an exception for the empty language ($\emptyset$) and the universal language ($\Sigma^*$).  For purposes of gaining a rough conceptual understanding, this technical detail can probably be safely ignored.