MOSFET drive is with respect to the source. Hence why you see "Vgs" all over the datasheet. You've wired the driver to the drain, which, isn't going to work very well I'm afraid.
Also unclear why you're using an N-ch bootstrap type driver with a P-ch transistor. P-ch is everything upside down, flip all the signs, that's it. Whereas N-ch needs Vgs > Vgs(th) to turn on, P-ch needs Vgs < Vgs(th) (and threshold is -2.5 to -1V in this case). But your driver only supplies positive Vgs. So I'm not sure how you intend to use it successfully here.
Mind, even if you tie source to Vs (the driver's supply reference), it's still wrong under certain conditions -- most importantly undervoltage (UV on the block diagram), which turns off the gate when insufficient supply voltage is available.
Which, hmm... that's gotta be a typo. Notice how they have "UV" going to an OR gate, to R (reset). Presumably, UV going true, means insufficient voltage is available. Reset means, well, reset the Q output (goes false / low). But they show Q going to a complementary pair of MOSFETs, an inverter. So they're showing the output as default high. Which I'm sure wasn't intended and they actually meant a noninverting gate driver output, not the two-transistor inverter.
On the other hand, they also don't provide a truth table; the "UV" function is insufficiently documented. Maybe it really does go high when disabled? That would certainly be critical information when using N-ch MOSFETs!
Regarding supply current, read the conditions. They give a shutdown current, quiescent current, and operating current. The first two are static, EN = 0 or 1. The last is for a 500kHz switching frequency into a 1nF load, in other words a 12nC equivalent gate charge (Q = V*C, and Vcc = 12V is given).
If the transistor is changed to N-ch, a problem still remains.
The bootstrap supply, is dubious without a pair of transistors switching alternately like in an SMPS application where these drivers are typically used. You are depending on load current to pull Vs down towards GND, and thus charge Vb through the internal diode. You cannot run continuous output-high (100% duty cycle), because the high side supply will eventually bleed through its charge, until it hits UV and forces itself off briefly. Preferably you do this manually rather than waiting for it, so as to maintain adequate Vgs(on).
Alternately, a charge pump or isolated supply can be added, to supply current to the high side continuously.
Also, this application sounds awfully familiar?.. Aha, yeah, here it is. I notice you didn't add a reply to this comment:
https://www.eevblog.com/forum/projects/is-it-okay-to-use-an-ldo-at-lower-or-the-same-as-its-rated-output-voltage/msg3593544/#msg3593544but also you seem to have ignored their advice... it is precisely what you need for a P-ch MOSFET, or various improvements for PWM capability.
Or move the load to the +V side (low side switched), so a N-ch is needed and only a regular (non-bootstrap) driver.
Tim
. You might see me again asking if my attempt at making a driver is correct 