Shortcircuit protection with PNP
This circuit doesn't protect anything. The transistor only adds to the stress on the power supply by turning on, and providing one more path for current to flow.

If the power supply is an ideal voltage source, then the transistor does not reduce the current which flows to the shorted output. The only way this circuit reduces the current is by adding to the load, so that the power supply's voltage sags more.

By itself it doesn't regulate in any way.

However, suppose that the input side is driven by a regulated power supply which already features overcurrent protection.  Then a circuit similar to this one can be used as a "crowbar" to deliberately trip the mechanism long before the device draws that much current. Say the power supply shuts down at 1A (or limits current to 1A), but you want that limiting or shutdown to happen when the load device draws only 100 mA.  Then you can pick the sensing resistor so that the transistor fully turns on at 100 mA (around 7 ohms).  Do not have any resistance in the collector circuit, so that the transistor shorts the power supply.

There is a better ways to use a PNP transistor to limit current to 100 mA. Rather than shorting out a power supply to trigger its limiting/shutdown, we can simply build a current limiter:



Here, the diodes are forward biased, establishing about a 1.4V drop. Diodes are used because their voltage will not vary significantly with a varying IN voltage, providing a stable reference. The transistor's BE junction needs 0.7V to turn on, leaving the remaining 0.7V across R1. We can choose the emitter resistor R1 to limit  the OUT current to 100 mA using \$R = V/I = 0.7/0.1 = 7\Omega\$.  

The value of R2 is much larger: it has to only conduct enough for the transistor to have enough base current drive and let some current through the diodes.

In this circuit, if OUT is short circuited to ground, still only 100 mA will flow (for the 7 ohm value of R1). In that case, the transistor will have some voltage across it, and will be dissipating heat. A power transistor with a heat sink may be needed. For instance, suppose the \$V_{CE}\$ voltage is 10V when there is a short circuit. In that case, the transistor is dissipating 1W.

You can use a single Zener diode instead of the two silicon diodes. R1 has to be adjusted then to keep the same current, in light of the Zener voltage.