Under 100 mA through the PIN diode, the 100 Ohm resistor will dissipate ~ one watt but the NPN transistor will have to dissipate more.
The full voltage from +85V to -5V is 90v, across the 100 Ohm resistor , 10 V; then I estimate almost 80V should be across the NPN transistor, close to 8W will have to be dissipated.
The requester never mentioned the PIN diode type; from the 100 mA open and 90V closed I guess it is used in a high-power switch application, the high power referred to the RF signal to be switched. For a RF power under one watt, the PIN diode current and voltage may be smaller than 0.1 A and 90 V.