You should have been able to quickly spot this:
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/8e4Bc3TK.png)
Note that there are very strong hints there. There are three current sources already identified for you. And they are all tied into the same diodes + resistor structure. So it's practically given away.
One diode accounts for the base-emitter diode drop of each of the connected PNPs. The remaining diode accounts for the drop across each of the emitter resistors, leading to a current source value. The author obviously is taking it as granted that the diode drop for the 1N4148, given the current yielded by \$R_{25}\$, will be about \$600\:\text{mV}\$. (I think this is a little low, given that about \$1\:\text{mA}\$ will be present in the two diodes.) So it's really easy to calculate \$\frac{600\:\text{mV}}{R_{23}}=4\:\text{mA}\$, \$\frac{600\:\text{mV}}{R_{29}}=6\:\text{mA}\$, \$\frac{600\:\text{mV}}{R_{31}}=6\:\text{mA}\$.
This tells you the entire function and design details of the entire upper part of the schematic. And it's not hard.
Also, since the only connection from this upper section and the lower section are via collectors, the lower section is free to float with respect to this upper section. This makes it relatively independent of the supply rail values with the only real change being some slight variation in source current values. So the lower section can be treated separately for analysis, with only the current source values showing.
With only \$6\:\text{mA}\$ sourced by \$Q_{22}\$ this circuit can't drive much of a load. It must be substantially more than \$2\:\text{k}\Omega\$ to have any hope of the stated \$+20\:\text{dBu}\$. So this is just a test circuit to evaluate different ideas. Not one likely to be used for most headphones or speakers.
This is also a class-A amplifier. You've only to look at \$Q_{23}\$'s arrangement to be certain about that.
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/khAENRb8.png)
As seen above from the global NFB circuit there, the gain should be \$A_v\approx 23\$. It'll be a little less due to the finite open loop gain of the system. But it should be close.
In your example in your question, however, the gain will be less with \$A_v\approx 11\$ (again, perhaps a little less than that due to the finite open loop gain, but very close I expect.)