Electrical – Do we have a voltage drop across this ideal current source

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.

A supply of 10 volts across 1k means 10mA is the maximum current you can get thru the MOSFET. That is the problem with your experiment - no matter how much more you turned the juice up on the gate, the aiming point for drain current is 10mA and can not be exceded no matter how hard you try.

Of course Vds decreased - the mosfet was turning really hard-on and acting as a very small value resistor. This small resistance forms a potential divider with the 1k to make a very small voltage that is nowhere near 10V.