Question: I understand when we say current, we mean charge (protons/electrons) passing past a point per second. And the charges have energy due to the e.m.f. of the power supply.

Now tell me, if a lamp has resistance and you hook it in the circuit, how will the current stay the same? The charges obviously lose energy in the lamp and so become SLOWER, which should mean current decreases, right?

[Edit]
All answers explained a bit of everything, so it was hard to choose one. If YOU are looking for an answer, please check the others too, in case the accepted one doesn't answer your question.

Answer: 
The question 

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> The charges obviously lose energy in the lamp and so become SLOWER, which should mean current decreases, right? 
> 
> 
> 

shows that you have a misconception about the motion of the conduction electrons. 

If you were correct then to maintain the same current around a circuit by a miracle more conduction electrons would need to contribute to the conduction process as you went around the circuit.  

If this did not happen then the conduction electrons would move slower and slower and . . . . . eventually stop? 

In fact what happens is that the electrons gain kinetic energy (and lose electric potential energy) between collision with the lattice (bound) ions from the electric field in the wire and then lose that extra kinetic energy to the lattice ions upon collision with them.  

The net effect is that the temperature of the material increases as the internal kinetic energy of the material has increased (the lattice ions vibrate more) and the conduction electrons move along the conductor with a constant average speed.
Your question is very nice, I have concluded that your question is how does the current remains the same when it supplies the power to the lamp for glowing it up.

There is an experimental law called [**Ohm's Law**](https://youtu.be/5AMTK7MEU_8), it's mathematical statement is $$ V = I R$$

Now, let's consider this circuit
[![enter image description here](https://i.stack.imgur.com/IqZKJ.jpg)](https://i.stack.imgur.com/IqZKJ.jpg)

If we consider that electrons flow from negative terminal to positive terminal and ignore the Quantum Mechanical aspects then for wire and bulb in this circuit we can write **Ohm's Law** like this $$ V\_{wire} = I\_{wire} R\_{wire}$$
$$V\_{bulb}= I\_{bulb} R\_{bulb}$$ .

we usually use the conducting wires whose resistance is small so from Ohm's law we can at once see that at any two points the voltage is smaller, but the resistance of the bulb is obviously much higher than the wire, therefore, the voltage between the two ends of the bulb is quite high. You should see that I have not used the fact here that current is constant, I have just used the experimental fact that voltage is directly proportional to resistance.

The relation of [drift velocity](https://opentextbc.ca/physicstestbook2/chapter/current/) with current is $$ I \propto v\_d $$ and I want you to understand that when the voltage will be high the electrons will have more energy and therefore will have high drift velocity, higher the drift velocity higher will be the current. Therefore, the current in the bulb will be higher than the current in the wire.

So, we have found that the current will be higher in the bulb and the electrons with high drift velocity collides with the atoms of the bulb, passing some of their energy to them and this energy is released as photons, now our *actually* traveling electrons have less energy (as they lost some energy during their collision with atoms inside the bulb material) so we can say that their drift velocity has reduced so the current and hence qualitatively we can say the current has now reached the same magnitude as it wass in the wire.

If anything is unclear you're welcome to ask it in comments.