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I tried in my design to make a common emitter stage amplifier using a pnp transistor and a npn transistor as a current source but I didn't know to bias either of them (well as for Q54 I think I've biased correctly I'm not sure) the simulation showed an operating point way lower than I expected (as you notice it's 1.178v) so the signal came out clipped. How can I increase my operating point value so it doesn't clip my signal?
In the place of Q55, if you had a resistor, you can easily fix the operating point by adjusting the value of the resistor but, it will reduce the gain. If you want to use Q55, you'll need to have a feedback loop to set the DC operating point and the bandwidth of that loop should be much lower than the signal frequency.
By saying Q59, Q58 in the title, you probably mean Q54, Q55 (Q1, Q2 in the schematics below). It seems you want to oppose Q1 to Q2 by connecting their collectors and applying the input voltage to Q1 and a constant input voltage to Q2. Such a simpler configuration (than the discussed in your previous question) is a common-emitter amplifier (Q1) with a simple active load (Q2) acting as a constant current source.
simulate this circuit – Schematic created using CircuitLab
There is only one main problem left to solve - how to apply the grounded input source voltage to the base of Q1 whose emitter is connected to Vcc and not to ground.
This problem is usually solved with a current mirror (see, for example, "current-feedback amplifier" in Google).
In the more complex configuration the input voltage of Q2 varies oppositely (differentially) to Q1's input voltage. It can be used as another input; thus we obtain a differential stage with dynamic load.
To bias a transistor means to add a constant voltage Vbias about 0.65 V. Figuratively speaking, we can do it in two ways:
by "lifting" the base with Vbias
by "lowering" the emitter with Vbias
I have already tacitly used the first option above by initially increasing the input voltages with Vbias. However, this assumes that the emitters are of fixed voltage (grounded). This means that the amplifier is "purely differential" - its two input voltages cannot change simultaneously within wide limits (the so-called common mode). So we have to "pull down" the common emitters with some "soft" element.
Collector-resistor stage: The humble resistor Re can serve as such an element as it faithfully served 100 years ago in the classic long-tailed pair. From one side, it is useful that by changing its resistance, the voltage drop across Rc varies, and this way we can set the quiescent output voltage; but from the other side, this is a disadvantage because the output voltage changes when both input voltages change (common mode).
By sweeping Re, we can precisely see how it determines Vout.
Dynamic-load stage: Let's now check if this is also the case with our dynamic-load stage. Surprisingly for us, Vout hardly changes when we change Re!
We can see it very well if we sweep Re.
Aha... the explanation is simple. As we change Re, both branch currents change. In the right branch, transistors Q1 and Q2 change their "resistances" in opposite directions; as a result, their midpoint does not change its Vout voltage (sort of like a virtual ground).
Collector-resistor stage: To improve the classic long-tailed pair, they replaced the emitter resistor with a constant current source Ie.
The results are the same.
Dynamic-load stage: As above, when changing Ie, Vout hardly changes for the same reasons.
This gives us reason to conclude that the emitter current source is not absolutely necessary in an amplifier stage with a dynamic load.
