Old version: 5Mhz processor and 8-bit bus.
New version: 20Mhz processor and 32-bit bus.
Processor speedup: $\frac{20}{5}=4$
Memory access speedup: $\frac{32}{8}=4$
Both of them would increase the speed by a factor of 4, but a process either uses CPU or uses the bus (for I/O) at a given time — not both simultaneously. Hence, speedup would be 4, and not 16.
Option B
Analogy: If glasses help you see up to extra 40 feet clearly, using glasses and a hearing-aid would still help you see up to extra 40 feet clearly. :P