I think the phenol will react much more quickly than the amine, especially under basic enough conditions to deprotonate the phenol. Its common enough for people to use tertiary amines as bases for all sorts of nucleophilic reactions without worrying about quaternary amine products. Typically to make quaternary amines from tertiary amines you have to do things like heat it vigorously (like 120C) for a long time, or use a halide scavenger like AgNO3 to drive it forward, since the reaction of the halide ion with the carbon alpha to the quaternary amine is also a favorable reaction. NO3 is a crappy nucleophile, and AgCl precipitates out.
I don't think the acylation reaction you drew would work. Under acidic conditions that amine has no lone pairs and would be an abysmal. nucleophile. Also, using strong base like aqueous NaOH to deprotonate the phenol probably also would deprotect your amine.
Give plain old Williamson ether conditions a try, finely ground K2CO3, alkyl halide, phenol, polar aprotic solvent.
Oh thanks for the tip regarding the high temperatures for forming quaternary amine ions! Good to know. And I think this can pretty much all be done at room temperature overall.
For the Williamson ether synthesis conditions, while K2CO3 be a strong enough base to deprotonate an aromatic OH? Also, during the deprotonation, won't the ammonium ion get deprotonated, leaving the lone pair exposed, and - in fact - sitting right on the other side of all the steric junk around the nitrogen, so it would be wide open to nucleophilic attack.
For the solvent I'm guessing just anhydrous acetone (Assuming the deprotonated compound is actually soluble in it), I guess I'm just trying to avoid something like DMSO because it has an annoyingly high boiling point and - for whatever reason - in always attracts swarms of flies into my workspace!
(Just as a side question since you mentioned K2CO3, I had some shipped to me and upon comparing how alkaline a solution of K2CO3 is compared to Na2CO3, I was shocked to see that K2CO3 was orders of magnitude more caustinc than the Na counterpart, so I just concluded that it must have been mislabeled... But is K2CO3 actually that much of a stronger base than Na2CO3? I usually look at K and Na as being more or less interchangeable)