Why are there three transistor pairs in Class B Power Amplifier circuit?
Why three pair of transistors are used in this?


I see only two pairs, the two Darlington pairs. The 2N3904 and 2N3906 are for output current limiting. These two are normally off unless so much current is flowing through the 1.2 Ohm resistors that enough voltage develops across them that the 2N3904 and 2N3906 activate. That will happen around \$V_{BE}\$ = 0.6 V. So around 0.5 A needs to flow through the 1.2 ohm resistors.


  Also, why there is unequal number of diodes


Note that the bottom diode is drawn as one diode but the text says: "3x 1N4005". I guess they mean 3 diodes in series.  I think that is plain wrong to draw it like that as it does not make it clear that the diodes should be in series. The reason to have 3 diodes instead of 2 diodes might be that the Darlington pair at the bottom (2N3906 + TIP32C) needs a higher biasing voltage compared to the Darlington pair at the top.

Edit the "3x 1N4005" appears to refer to all three diodes that are drawn. That would mean there is indeed only one diode at the bottom side. That means that the bottom PNP Darlington pair is biased at zero current and will only "kick in" when there is some signal. My guess is that this amplifier will have significant crossover distortion as a result of that.

In a standard class B amplifier it is typical to have the same number of diodes as there are Base-Emitter voltages stacked in the transistors. Here there are two Base-Emitter voltages stacked at both sides. Adding another diodes increases the voltage across the emitter resistor (that's the 1.2 ohm resistor) and increases the biasing (also called quiescent) current. If you do that on top and bottom sides the amplifier becomes a class AB amplifier.