How protective diode protects transistor from breakdown?
The diode is in place to protect the transistor from reverse \$V_{BE}\$ breakdown. If you reverse-bias the base-emitter junction by taking the transistor's emitter to ~\$6\rm{V}\$ more positive than the base, it will breakdown and begin conducting.

This reverse breakdown damages the base-emitter junction, causing a degradation in \$h_{FE}\$.

By placing a diode as shown, the reverse-bias voltage is limited to ~\$0.7\rm{V}\$; attempts to apply more voltage will be futile, since the diode will conduct a lot of current and prevent increased voltage. This protects the transistor's base-emitter junction.