Active multiplier vs step recovery diodes

What is the main advantage of active multiplier(HEMT based/MESFET BASED) over the SRD type multiplier? what are the limitations of each type! plz guide me .

The active multiplier will produce mostly the signal you want. The diode produces numerous harmonically related signals that you will have to filter out.

Another advantage is a greater output signal. As the active multiplier includes amplification. With an input signal of 2dbm you may get 15dbm.
A SRD will generate harmonics with lower amplitudes as the order of them goes up. You must select the correct one with a filter that will attenuate more. So now you have a conversion lost instead of gain. Requiring amplification.

Varactor-diode frequency multipliers offer a good efficiency but are tricky to design. Also, they mostly operate at a single frequency input and output. They are very sensitive to input power.

Active multipliers are easier to use but more expensive. The can cover a wide frequency band and are not sensitive to input power variation.

since active multipliers have the driver amp (and possibly output amp)stages built in, they are easy to use. Just drop them on your PCb and you are multiplying away. But in my experience, the are a little touchy as far as stability goes, so you have to experiment around before you are sure they are happy with that bandpass filter you are adding. Also, the phase noise will not be as good as schottky or varactor multipliers.

SRDs require a high drive power, so you are wasting DC power in a driver amp. It might be a little harder to filter the output ofs! a varactor diode multiplier, as all harmonics are present, while in an active multiplier you likely have only even mode or odd mode harmonics. And logistically--good luck finding a good SRD nowadays

What will be the efficiency of Active doubler circuit(x2) operating at Class-B region. Is it 78% just as in normal we are telling in terms of amplifies!Plz tell as i ahve to face an interview on 6th of this month!

It seems a little strange to ask the efficiency of doubler.
But you can calculate as the following:
Class B Amplifier's efficency is 78% for the base wave. So say 2H is about -20dbc lower than base wave, so 2H energy is 100 times less than base wave, that is 0.78%.