Electronic – the point / name of this circuit? Reverse voltage protection + current mirror

To figure out \$V_{GS}\$, all you need is the MOSFET current equation. You have all the parameters you need for that, and you did the calculation correctly:

$$I_D = \frac12 k' \frac WL (V_{GS} - V_t)^2$$

$$250 \mathrm{\mu A} = \frac12 \left(80 \mathrm{\frac{\mu A}{V^2}}\right)(3)(V_{GS} - 1 \mathrm V)^2$$

$$V_{GS} \approx 2.44 \mathrm V$$

Now, if you want, you can use KVL to figure out the other voltages in the left branch of the circuit:

^{simulate this circuit – Schematic created using CircuitLab}

$$5\mathrm{V} - V_{C} - V_{GS} - -5\mathrm{V} = 0$$

where \$V_C\$ is the voltage across the current source. You can plug in your \$V_{GS}\$ to get:

$$5\mathrm{V} - V_C - 2.44\mathrm{V} - -5\mathrm{V} = 0$$

$$V_C \approx 7.56\mathrm{V}$$

Normally you don't need to figure out the voltage across the current source for a homework problem. (In real life, you need to make sure you have enough voltage, especially for more complex circuits like a Wilson current mirror.) You might have to find the voltage across the output transistor, since that will let you take channel length modulation into account and determine the error of the mirror.

Basic function of current mirror here is to "copy" the Ibias current to main circuit via dividing by 25 (due to transistor sizes 600 vs. 24).

In order to maintain stability of circuit, we generally use transistors in saturation. Very small current may cause transistor not to saturate.

We can use transistors to obtain Ibias instead of resistors. If you maintain saturation of transistor (see diode connected), you can get stable and constant current.