Diodes in LDO schematic

You have 500 mV headroom, so a LDO should be possible. A diode is not a exact voltage drop. The output of a LDO will be much better regulated. Unless this is a very high volume product and the extra few cents actually matters, use a LDO.

The input voltage will need to exceed the output by the Vbe of the transistor + VCE(sat) + V(dropout)[LDO]. Under most conditions, that is about 1V plus the LDO dropout voltage. If you can live with the added V(in), then the outboard boost will operate.

The boost circuit will turn on at about 0.6V/R1. If you wanted to engage the boost at 30mA, then the resistor is 0.6/0.03 = 20 ohms. That 30 mA will (mostly) still pass through the regulator. The mostly refers to the fact that the transistor base current must be supplied from the input, so the output provided from the LDO is 30mA - Ib. All other current is being supplied from the outboard boost transistor.

The actual voltage at which the boost engages at will vary significantly over temperature due to the -2.1mV/Kelvin temperature coefficient of the base-emitter junction.

The actual output current is load dependent. As this circuit has no current limit, you will need to ensure that the transistor can handle the expected load current.

Make sure you choose a resistor with a suitable power rating (in this case 30mW minimum - I would choose a 100mW device at a minimum for this function).

HTH