Electronic – Why is a diode-connected BJT more “ideal” than a simple pn diode

It is there to keep the transistor's current less susceptible to temperature changes.

In the case of Q1:

Suppose that instead of having R1 and D1, Q1base was connected directly to ground.
Emitter current would be: $$ I_{e} = \frac{20V - V_{be}}{R_{2}}$$

You can see Ie is susceptible to variations in Vbe, which has a known dependency on temperature (T), so you might as well express it as: $$ I_{e}(T) = \frac{20V - V_{be}(T)}{R_{2}}$$

But with the diode, if they are matched and thermally bonded: $$ V_{diode}(T) = V_{be}(T) $$

So now: $$ I_{e} = \frac{20V+V_{diode}(T)-V_{be}(T)}{R_{2}} $$ Which simplifies to: $$ I_{e} = \frac{20V}{R_{2}} $$ Independent of Vbe, and its variations with temperature.

The diode is effectively providing the little voltage offset that would be needed to compensate for Vbe changes with T, in order to maintain a constant current.

What

The voltage drop is a fundamental characteristic of forward-biased doped-silicon semiconductor junctions. There are diodes with a smaller voltage drop but all diodes will have a voltage drop.

You can look at a graph of diode characteristics but all this tells you is that Vd exists, not why

Why

To understand this in detail, you need to read about semiconductor physics. See PN junction voltage drop

The way I think of it is as follows:

Because of the doping of the silicon, "free" electrons diffuse from the n-type region adjacent to the junction. These electrons combine with holes in the p-type region. This leaves a "depletion-zone" that lacks carriers and is therefore insulating. To drive current through that region you have to apply an electric field to create a force that moves carriers into the depletion region. This electric field across the junction is what we measure as a voltage across the junction.

^{From Wikipedia}