Balanced Class A Quiescent Current Regulator (OPS of Amplifier)
The (AC) signal from VA does directly end up at the base of both Q3 and Q4 through the capacitor CB.

The DC voltage at the base of Q3 and the base of Q4 cannot be the same so we cannot short Q3 and Q4's bases. If we did that then the whole DC biasing of the circuit would be influenced and it would not work anymore. So we cannot short the bases for DC. But we can still short them for AC (the signal), that's what capacitor CB does.

The capacitor CB and the rest of the circuit act like a DC voltage source (like a battery) so the signal at VA also appears at the base of Q3 but with an additional DC-voltage added.

How much is that DC voltage?

Follow this path: Base-Emitter of Q3 + BE of Q1 + V(RE1) + V(RE2) + EB Q2 + EB Q4  so about 4 x a Vbe + the voltage across RE1 and RE2 which depends on their value and the DC biasing current through them.

Sure but how much voltage will there be across RE1 and RE2?

That's where Q5 and Q6 come in, if the voltage across RE1 and RE2 starts to become more than 2 x 0.6 = 1.2 V then Q5 and Q6 start to conduct, they then "steal" some of the base current of Q3 and Q4 making Q3 and Q4 conduct less which makes Q1 and Q2 conduct less which makes less current flow through RE1 and RE2.