If the Vcc voltage for the 74HC238 is 5V it might make sense to keep the resistor R18, if Vcc is 3.3V you don't need it (and it's better without it). There is enough output Z in the 74HC chip to avoid emitter follower stability issues.
I don't see any point in using a 300V transistor an a 3.3V circuit. It's more expensive than something like a 2N4401 and has much poorer performance. Gain is as follows:
So it could be as low as 15 or so at low temperature and 56mA (using worst case gain of 25 at 25°C/50mA and adjusting for temperature and higher current), meaning the base will be drawing 3.7mA. That would drop 3V across the 820 ohm resistor, which is obviously going to mean the LED current will be mostly determined by the transistor gain. This is not what a good design does- gain should ideally have no effect on the operation.
So, assuming your Vcc is 3.3V throughout- get rid of R18 and use a 2N4401.
Keep in mind that the transistor will still drop about 0.7V plus whatever the 74HC output drops, so hopefully your LEDs are low enough Vf that such a drop is acceptable. For reliable operation, that probably means a red LED only.
If your 74HC is running from +5 you will need the base resistor and you can saturate the transistor so all is well.
The way to do it with little drop is to use a PNP transistor (or p-channel MOSFET), but your control signal will be inverted so probably the HC238 is not what you want. If your LEDs are blue or white, especially, you're probably marginal even with that.
P.S. The 10K is probably not necessary unless you are doing something else with the digit scan signals. Unless you are worried about leakage at very high ambients and slight LED visibility in a very dark room.