How does soft start circuit work in UC3842B?
but how soft-start circuit controls comp pin voltage?

The comp pin is pin 1 which is connected to the output of the Error Amplifier EA.

Both soft start circuits simply pull down the EA's output. sure they have drawn the EA as if it is an opamp so you would expect that pulling down the output would have no effect as an opamp (used with feedback) will simply deliver more current so the voltage would not change.

However I think that the Error Amplifier is not really an opamp. I think it is an OTA (Operational Transimpedance Amplifier), it is like an opamp but instead of having a voltage output it has a current output.

What makes me conclude that?

Well, notice the 1 mA current source which is connected to the EA's output. If the EA was an opamp, that would be pointless. But when the EA is an OTA then it does make sense to do this.

Suppose the input voltage difference of the EA is zero (no error) then the EA will output no current. A 1 mA DC current is added at the output which will flow into the diode to the right. At startup circuit the soft-start circuit can subtract current making the current into the diodes even smaller. That then affects the PWM signal at the output.

Depending on the soft start time you need you could choose the first or second circuit. For long startup times you might need a large capacitor, then the second circuit might be better (the one with the PNP) as it will make the PNP deliver the actual current so the capacitor can have a smaller value but still achieve the same startup time as the first circuit.