High-side driver IC use of diode in the npn transistor BE
Call the bipolar NPN Q1 and the base emitter diode D1.

The diode replaces the lower transistor in what would otherwise be a 2 transistor driver. It provides a pull down path using the previous driver stage.

Usually the gate of M2 would be driven by a 2 transistor driver which was able to pull its gate high to turn it on and to pull its gate low to turn it off.
The designer has decided that M1 has enough drive capability to be able to be used to pull M2's gate low. M1-drain is connected to M2-gate by D1. When M1 is on it pulls M2-gate low via D1 and turns M2 off.
To turn M2 on the deigner COULD have used just R to provide pullup drive to M2-gate. However, R cannot supply enough current to provide the gate charge required by M2 in an acceptably short time. So Q1 is provided as an emitter follower to pull M2-gate high when required.  

In this manner M2 is provided with bipolar drive - high via Q1 and low via D1 + M1.

Note that this answer is another way of saying what Wouter's answer says.