The complement of a graph is what you get when you replace (a) all the edges with non-edges, and (b) all the non-edges with edges.
So, here's the 4-node path and its complement:
The complement also happens to be a path.
The complement of a path on 2 nodes is the null graph on 2 nodes (i.e., 2 nodes, no edges), drawn below. This would generally not be regarded as a path, so this is an error in the statement given.
If the single vertex graph is considered a path, then its complement is also a path.
So probably the statement should be:
There are only two paths such that their complements are also paths: the path on 1 node, and the path on 4 nodes.
Oh, and the proof is essentially: (a) check small cases, then (b) for $n \geq 5$ nodes, the complement has a vertex of degree $ \geq 3$, so cannot be a path.