Just a note: the previous answers are ok, however you're not too far from the correct trivial reduction:
if $\sf{P} = \sf{NP}$ then any $L \in \sf{NP}$ is Karp reducible to the language $\{1\}$ (just map in polynomial time every $x \in L$ to 1, every $x \notin L$ to 0), which is trivially a sparse language
The converse direction: "if an $\sf{NP}$ complete language is Karp reducible to a sparse set then $\sf{P}=\sf{NP}$" is certainly more interesting and it is known as the Mahaney's theorem:
Let $c$ be a constant and $A$ be set such that for all $n$, $A$ has at most $n^c$ strings of length $n$. If $A$ is $\sf{NP}$-complete then $\sf{P}=\sf{NP}$.