Eu encontrei esta solução. A única falha realmente é que requer a seleção de um caractere especial não presente nas listas em questão.
EMPTY :=
SPACE := $(EMPTY) $(EMPTY)
# Pick a non-space character that NEVER appears in any word in $S. Here we use a comma.
XX := ,
S := This is very funny world and YOU is the funniest of all!
FIND := YOU
$(info Find $(FIND) in $(S))
S := $(XX)$(subst $(SPACE),$(XX),$(S))$(XX)
$(info S is $(S))
S := $(subst $(XX)$(FIND)$(XX),$(XX)$(FIND)$(SPACE),$(S))
$(info S is $(S))
S := $(firstword $(S))
$(info S is $(S))
S := $(subst $(XX),$(SPACE),$(S))
$(info S is $(S))
IDX = $(words $(S))
$(info IDX of $(FIND) is $(IDX))
# Combine into a giant function that can be called.
IDX = $(words $(subst $(XX),$(SPACE),$(firstword $(subst $(XX)$(2)$(XX),$(XX)$(2)$(SPACE),$(XX)$(subst $(SPACE),$(XX),$(1))$(XX)))))
$(info Index of $(FIND) is $(call IDX,$(S),$(FIND)))
# Define a function that returns the list element of the first argument that
# is at the same position as the second argument is in the third argument.
ATIDX = $(word $(call IDX,$3,$2),$1)
Assim, a regra seria parecida com:
%.a: %.b
myprogram $(call, ATIDX,$(Y),$(*),$(X)) $(call ATIDX,$(Z),$(*),$(X)) $< $@