Yes, it does matter, and (A) is the wrong order for a switching application.
The (B) circuit is the only one where the transistor acts a saturated switch, as you likely intended. It's unlikely to be damaged as long as you don't exceed the transistor's maximum collector current and have enough base drive current available to keep it saturated.
In the (A) circuit, the transistor is in the linear region, and will be dissipating quite a bit of power. Heavier loads may destroy it, and you won't be able to fully turn such loads on anyway. The well-being of the transistor requires investigation of the SOA - Safe Operating Area - graph in the datasheet.
It happens that on this very website you already have all the tools to answer your own question!
Let's set up both circuits side-by-side. Don't forget about the base resistor that you've omitted from your diagram!
simulate this circuit – Schematic created using CircuitLab
We can now plot the LED current vs. the value of the series resistor, and compare the behavior of both variants, as the load resistance is swept from 50 Ohms to 10kOhms.
As you can see, if you want to drive the load harder - perhaps it's not an LED but, say, a solenoid, or a motor - the circuit (A) "runs out of oomph" and doesn't fully turn the load on.
That's because a transistor is controlled by the base-emitter voltage. In a circuit where emitter is not at a fixed potential, but you drive the base with a fixed voltage, you'll be at the mercy of the load's behavior and won't be fully in control of the transistor.
But so what, the transistor is still switching, right? Well, yes, in circuit (B) the transistor always acts as a switch. But in circuit (A) it only acts as a switch when switching relatively light loads. As the load gets heavier, the transistor stops acting as a switch, and acts like an additional series resistor.
To see this in action, let's plot the voltage across the transistor in both variants of the circuit, as we sweep the load resistance from 50 Ohms to 10kOhms.
Observe that the transistor in the (B) variant acts like a saturated switch, with a fairly low voltage across it.
In the (A) variant, the transistor does not work in the saturated switching region, and the voltage drop on the transistor can be arbitrarily large. As the load current increases and the collector-emitter voltage grows, eventually these two parameters will get outside of the safe operating area (SOA) and the transistor will be destroyed.
To see this, let's plot the power dissipated in the transistor in each of the circuit variants, as the load resistance is swept from 50 Ohms to 10 kOhms:
The transistor in the (A) circuit dissipates up to two orders of magnitude more power than in the (B) circuit. We can see this by plotting the ratio of the power dissipated by Q1 to power dissipated by Q2, as the resistance is swept:
Is there a way to make the (A) circuit operate as a saturated switch? Sure, just use a transistor whose emitter can be connected to the fixed VCC potential: a PNP transistor.
simulate this circuit
Note that while the NPN transistor's control voltage is referenced to ground, the PNP transistor's control voltage is referenced to VCC. When the base of the PNP is at emitter potential VCC, the PNP is off, just as when the base of the NPN is at the emitter potential GND.
Both transistors drive similar currents into the loads across the entire range of load resistances:
They both also operate as saturated switches, with only a small saturation voltage present across the transistor:
All the plots were generated using CircuitLab, and the DC Sweep simulation parameters are pre-entered, so you can re-run the simulations easily from each of the two circuits. You'll get all the plots I've shown above for their respective circuits.