YES

1 decompositions
 #0 -----------
   1: f(x) -> g(f(x))
   2: h(x) -> p(h(x))
   3: f(x) -> h(f(x))
   4: g(x) -> p(p(h(x)))

@Rule Labeling
--- R
   1: f(x) -> g(f(x))
   2: h(x) -> p(h(x))
   3: f(x) -> h(f(x))
   4: g(x) -> p(p(h(x)))

--- S
   1: f(x) -> g(f(x))
   2: h(x) -> p(h(x))
   3: f(x) -> h(f(x))
   4: g(x) -> p(p(h(x)))