YES

1 decompositions
 #0 -----------
   1: f(x) -> g(k(x))
   2: f(x) -> a()
   3: g(x) -> a()
   4: k(a()) -> k(k(a()))

@Strongly Commuting
--- R
   1: f(x) -> g(k(x))
   2: f(x) -> a()
   3: g(x) -> a()
   4: k(a()) -> k(k(a()))

--- S
   1: f(x) -> g(k(x))
   2: f(x) -> a()
   3: g(x) -> a()
   4: k(a()) -> k(k(a()))