So at 100 V input, there will be 23 V across the load and77 V across the transistors? At any load current above 25 mA you will need heatsinks and possibly a fan.
Back to the voltage "drop"...
(1.4v is quite a % at 8v).
That would matter if that 1.4 V were in series with the load, but it is not. The *saturation voltage* Vcesat (not Vbe) is in series with the load, and by definition that is lower than Vbe. The only "efficiency" difference between a MOSFET and a BJT is the gate energy versus the base-emitter energy. A power MOSFET typically has a higher gate voltage for full enhancement than a BJT base-emitter voltage does for saturation, but the steady-state gate current is so much less than the base current that the total energy is less. But even a logic level MOSFET needs over 2 V for full enhancement, while you state that the BJT needs only 1.4 V for saturation.
Efficiency is the output power divided by the total circuit power. At its heart, the circuit is a linear regulator, so by definition it is the least efficient power regulator topology possible, equivalent to using relays to switch resistors in series with the load. No matter how many power devices share the load current, the efficiency is the same. Since the load current is a secret, let's assume 1 A. At 100 Vdc input, there is 23 W dissipated in the (secret) load and 77 W dissipated in the regulator. This is almost completely independent of the number or types of regulator pass devices. Almost, because (not counting the current through the various zeners) if the only pass element is a power darlington, its base current is something less than 1 mA while the gate current of an equivalent MOSFET will be under 1 uA. That is a thousand-to-one difference, but in something that is very small to begin with. The base current is less than 0.1% of the total circuit current, so it is a barely measurable effect on efficiency.
What I *think* you are trying to say is that a fully "on" power MOSFET has a lower Vds voltage drop in series with your load than would a fully saturated BJT Vce. That usually is true; but not always, depending on the rest of the circuit parameters and the specific components chosen. To have the BJT conduct only when the input is above 24 V, put a 22 V zener in series with the base. This increases the transistor's effective "turn on voltage" to 23.4 V. Note that typical zener diodes are only 5% or 10% accurate, the zener voltage changes with its series current, and the BJT base-emitter conduction voltage also changes with both base and collector current.
A final note, many power darlington transistors have two base-emitter resistors built-in. These need to be a part of the overall circuit efficiency calculation.
Based on the information you have released so far, sharing the load current between two power devices is not necessary, and decreases both the overall circuit reliability and its "efficiency".
Also, I think what you are trying to build is a 23 V voltage regulator that has a very low series voltage drop when the input is less than 23 V. This is called an LDO (Low Drop Out) circuit. There are many examples on the internet, and control chips for this specific purpose if you don't want to grow the circuit from scratch.
What is the peak load current?
ak