Yes it's time delay switch using the transistor as the "switch"
On power up the 100R should be disconnected from the PT2399. The transistor should be turned off initially and that means there is no path to 0v through the transistor collector-emitter for the 100R, effectively switching it out of circuit. When the 47uF capacitor charges up sufficiently (the time delay), the transistor turns on because there is enough voltage to "forward bias" it's base-emitter junction. In your case, this appears to happen about 0.5V.

This was the easiest schematic to find - ignore the red modification notes.
Now, the time delay is how long it takes the 47uF to charge via the 68k resistor. This must be long enough for the PT2399 internals to get running normally with only the 2k2 resistor setting the delay time. Then the 100R can be switched in, making the PT2399 delay shorter to suit the chorus effect.
If you think about it, the time delay due to the 47uF capacitor only happens IF the capacitor is completely discharged before power is switched on. I'm not sure that this is always happening. Indeed the 47uF may keep a small charge for a very long time after 9v power is removed.
A cure might be to fit another 1N4148 diode in series with the transistor base. This would raise the time it takes for the capacitor charge to turn the transistor on AND negate the effect of any small charge remaining in the capacitor from the last time power was applied. I also wonder if it wouldn't be better for the existing 1N4148 diode that is meant to discharge the capacitor to have it's cathode connected to the 9v supply instead of 5v. I suspect it will be able to discharge the capacitor faster that way.
This is a bit technical, but MOS devices like the PT2399 tend to have very high resistance to the power supply once the supply voltage falls to about 1v, because none of the MOS transistors in it can be turned on at that low a voltage, so there isn't much resistive loading on the 5v supply that could discharge the 47uF capacitor below that 0.5v level. It could perhaps have 0.4v on it for quite some time.