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Here is the hybrid-\$\pi\$ model:

enter image description here

with \$I_c\$ and \$I_b\$ going in, and \$I_e\$ going out. The same circuit can be used for both NPN and PNP, the only thing changing being \$V_{be}\$ for NPN and \$V_{eb}\$ for PNP.

The currents direction does not change, which is what I don't understand. We know that for a PNP in active mode, \$i_c = i_e+i_b\$.

For a NPN in active mode, \$i_e = i_b+i_c\$, which corresponds to the current flow shown on the hybrid-\$\pi\$ model.

Why is it the same for PNP?

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It's a small signal AC model, so the DC currents don't matter. Since \$ \frac{dI_{C}}{dV_{BE}}\$ is the same polarity for both, we can use the same model (i.e. a larger b-e voltage results in a larger load current, just they are both negative for the PNP version (which equates to the same result as the NPN)

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\$+g_m\cdot v_{be} = -g_m\cdot v_{eb}\$

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Directions of \$i_b\$ and \$g_m v_{be}\$ are correlated. They both must go to the emitter or both go out of it.

If it is easier for you use this 'inverted' model:

pnp small signal model

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The + and - sign at the base-emitter side and the arrow at the collector-emitter side only refere to the reference direction, i.e. they do not make any statement about the actual current direction. They indicate only what direction is considered positive and what direction is considered negative.

In case of the NPN transistor:
Base: current is flowing into the base, i.e. direction is from + to -, i.e. base current is positive.
Collector: current is flowing in direction of the arrow, i.e. collector current is positive.

In case of the PNP transistor:
Base: current is flowing out of the base, i.e. direction is from - to +, i.e. base current is negtaive.
Collector: current is flowing opposite to the direction of the arrow, i.e. collector current is negative.

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