Two voltage sources on the same circuit
First of all: 9V means that you are using a 9V "transistor" battery (like this), right?

If so it won't last very long: I suggest you to use a different approach (e.g. four AA batteries in series).

As for your question, I couldn't find the power consumption for these leds. Try to measure it yourself, setting the max brightness and white color. I think that, anyway, the max current for each LED should be 60mA (20mA for each color). This means that each strand absorbs 1.2A (quite a lot!).

This power requires you to use at least a switching regulator. I always used some small modules I found on ebay (they have a LM2596 IC, so just search LM2596 on ebay to find them). They are rated for 3A, so you should put at least two of them (1.2 * 3 strands = 3.6A > 3A).

The connections are quite straightforward: the board has four pins: IN+, IN-, OUT+ and OUT-. IN+ of all the modules go to the positive terminal of the battery, IN- and OUT- of all the modules go to the common ground node (negative terminal of trinket, battery and leds), OUT+ is the 5V regulated output. Oh, by the way, they are adjustable modules, so BEFORE ATTACHING THE LEDS use a multimeter to set the correct voltage (if you attach the leds without regulating the voltage you can destroy them).

Since you will have a lot of current left on the second module, you can power the trinket from there (so you will not waste a lot of power).

One last thought: 9V batteries usually have a capacity of around 600mAh; if you use one of them with the switching modules and drain 4A from the 5V your battery will last 0.6Ah / 4A * 9/5 = 0.27h = 16 minutes; if you switch to 4 AA batteries (around 2000mAh) it will last 2Ah / 4A * 6/5 = 0.6h = 36 minutes. More than double (calculations made rounding up the power drained by 21 leds set at max brightness and white color)