I have this very simple solution using only one bjt.
With your window og 4 - 25 mA, you'll have to find out the voltage over the resistor. What value, let me guess 100 ohms. Lowest voltage must be U_R.low = I.low * R = 4mA * 100ohms = 400mV. The highest voltage is U_R.high = I.high * R = 20mA * 100ohms = 2V.
Because of the voltage follower, we don't care about the voltage drop from base-emitter, the opamp will handle that.
Then you'tt have to put your 5k potmeter in series with two resistors. One above and one belove. The potmeter will work in a voltage window of 2v - 0.4v = 1,6v. That means the current through R.var = 1.6V/5kΩ = 0.32mA.
Value of R2 must then be 0.4v/0.32mA = 1250Ω. Value of R1 must be: (24v-2V)/0.32mA = 68.75kΩ.
That means you can use your 1.2kΩ resistor get litle less then 4mA minimum, and you can use a 68k (or a 82k) resistor.