A constant current source simulation using transistor: problem in Proteus
Your simulator is poor in a couple of ways:


The ammeters only show tens of milliamps in the least significant digit. You need finer resolution on their readings.
The ammeters apparently don't round values up, but instead simply truncate. So a value of \$9\:\textrm{mA}\$ will read as \$0.00\:\textrm{A}\$ rather than \$0.01\:\textrm{A}\$, as you might normally expect.
The TIP122 is a Darlington. It is impossible that \$V_{CE}=0.02\:\textrm{V}\$ in the circumstances you show there. As low as 20 times that much, perhaps. But that figure isn't right for a TIP122.
The TIP122 is a Darlington. It is impossible for \$V_{BE}=5\:\textrm{V}-4.39\:\textrm{V}=610\:\textrm{mV}\$. Again, it might be as low as \$1.2\:\textrm{V}\$. But again, that figure isn't right for a TIP122.


As a result, I'm almost certain that the transistor shown in your diagram is NOT a TIP122, at all. Nor is it any existing Darlington. But something else. Probably some regular NPN BJT model is being applied here.

My conclusion is that your transistor is a simple NPN BJT and not a Darlington.

Let's analyze the following (much more likely) case:







simulate this circuit &ndash; Schematic created using CircuitLab


Here, the base is nailed to \$5\:\textrm{V}\$. This means the emitter is going to be about \$700\:\textrm{mV}\$ less, or about \$4.3\:\textrm{V}\$.
This means that the emitter current will be about \$I_E= \frac{V_E=4.3\:\textrm{V}}{R_1=430\:\Omega}\approx 10\:\textrm{mA}\$. Note that this is very close to what your ammeter says.
The collector cannot be lower in voltage than the emitter. It can get close. Your reading says \$0.02\:\textrm{V}\$, which is possible. This means the voltage across \$R_2\$ is \$100\:\textrm{V}-4.3\:\textrm{V}-0.02\:\textrm{V}\approx 95.68\:\textrm{V}\$. Note that this is very close to what your voltmeter says.
The collector current will then be \$I_C=\frac{95.68\:\textrm{V}}{R_2=50\:\textrm{k}\Omega}\approx 1.91\:\textrm{mA}\$. This is too low for your ammeter to read it out. So it shows zero, instead.
The base current will then be the difference, or in other words: \$I_B=I_E-I_C= 10\:\textrm{mA}-1.91\:\textrm{mA}\approx 8.09\:\textrm{mA}\$. This is also too small for your ammeters and is apparently truncated to zero in your reading.
Given the base current is so much more than the collector current, the BJT is in fact very heavily saturated. Which is consistent with the measured \$V_{CE}\$ in your diagram.


The analysis is complete. There are no mysterious \$500\:\textrm{V}\$ voltage drops anywhere in the circuit. The currents make sense, the voltages make sense, the heavily saturated condition of the NPN BJT makes sense.

The problems are: (1) The simulator's inability to round values, or show them with the needed precision here; (2) The fact that you aren't simulating a TIP122, but instead some regular NPN BJT; and (3) Your internal mental state which incorrectly concluded that the measured emitter current must also be taken as close to the collector current. Instead, the base current is quite a bit larger than the collector current, so most of the emitter current is accounted for by examining the base current instead.

That's all there is to it.