The max Vgs of your P-MOS is the maximum voltage between its source, and gate. As its a P-channel MOSFET, the 'source' will be at your Vin potential - 28V. So, as the max Vgs of that device is 20V. 28 - 10 = 8v. That is the lowest voltage you must provide to the gate.
I'm sure there is a known way of doing this, (without lowering your input voltage) but all I can tihnk of right now is to use a zener diode in series with drive signal (ground referenced) to limit the voltage to a minimum of say, 18V (Vgs of 10V) instead of 0v. Then using bipolar transistors as level shifters/inverters to shift up your gate drive from the open collector output. Like this but with a zener in series with the 100R resistor on the first transistor:
https://www.boerde.de/~mw/msp/HighSidePFET.jpg
The first transistor in that circuit being the open collector drive of the MC34063
Perhaps using a boost circuit used for high side N channel MOSFET's. Like this:
**broken link removed**
But with opposite polarity? (with the transistors, diodes, and supply in reverse.)
Consider either a device with a higher Vgs (unwise, as that likely increases gate driver requirements as far as I'm aware..) or perhaps use an N channel with a boost?
**broken link removed**
One more link for old times sake:
https://www.electro-tech-online.com/custompdfs/2011/09/AND8291-DPDF.pdf - page 3! Maybe use a 12 or 15V zener in place of the 6.2V one there.
That should give some food for thought