You have received a large amount of useful input related to the question that you asked and it will be of assistance to others who read these answers in future.
However, you have been wasting people's time and confusing your self because you stated what you thought you needed to do to fix your problem instead of telling your people what your problem is. While there is some overlap the answers that have been given mainly relate to things that you are not trying to do. While they do somewhat address what you are trying to do, the diagram that you provided would make almost no sense in most contexts and is NOT doing what it appears to be doing.
Lesson: "Tell us what you actually are trying to do and we will tell you the best way to do it".
Real question: See Maxim DS1822 Data Sheet -
PAGE 5 - POWERING THE DS1822 and
page 6 SUPPLYING THE PARASITE-POWERED DS1822 DURING TEMPERATURE
In the related diagram below Vpu is a "weak pullup" and the FET is a "strong pullup".
When the Vdd pin is grounded, power supply energy may be provided via the DQ line and is stored in an internal capacitor Cpp (C parasitic power). During most portions of operation the "parasitic" feed provides enough current Ipp at an acceptable voltage to power the IC. During some operation Ipp is inadequate and the iC must either be powered via Vdd or via a higher current source (see data sheet page 5). During these high current operations the FET is turned on to provide extra supply current. This low resistance power feed clamps the bus high and presents it being used for signalling by other ICs on the bus, so the 'strong pullup' is enabled only for as long a period as required.
SO:
You do need a FET for pullup, you do need a high side FET, this need is most simply met with a P channel FET - all as advised by others.
As Vmicrocontroller (Vmcu) is >= V1_wire_bus, the FET is not being used as a level converter but as a high side power supply switch.
Choosing a MOSFET:
Connecting a suitable P Channel MOSFET as shown in the diagram will fill the need.
Many FETs will do the job.
Rdson / On resistance: MOSFET must have low enough on resistance = Rdson for the task.
A MOSFET that dropped 0.1V at 2 mA would probably suffice
Rdson = Vdrop / Iload =
= 0.1v/2 mA = 50 Ohms.
You'd have immense difficulty buying a P Channel FET with Rds = 50 Ohms = normally available ones are typically 50 to 5000 times BETTER (lower Rdson) ie 1 Ohm down to say 10 milliOhms.ie ANY P Channel MOSFET that meets other specs will have an OK Rdson
Gate operating voltage = Vth or Vgsth:
Vth or Vgsth should be << Vcpu.
ie the μP (microprocessor) should easily drive the MOSFET.
A 3.3V μP will JUST operate a MOSFET where Vth = 3V.
Operation will be better at Vgsth = 2.5V
and better again at 2V. Lower again does not hurt.
Vds_max > say 10V is OK - 20V or 30V better. > 30V OK.
Ids_max is so low as to be met by anything.
The horrible BSS184 - datasheet here is 20 cents in 1's at Digikey and does the job well enough. Digikey and others have many more that will do a better job - but not needed here.
You will need to compare the current draw of the 5V regulator with a light load vs. the smaller 3.3V switching (or even linear) regulator. Chances are it will draw a fair bit more current. This is your trade-off to make. Remember that when the big print says 86% or whatever efficiency that is the highest efficiency point of operation. The efficiency will go to zero at zero output current and may be worse than even a linear regulator at relatively high output currents (less than ~50mA in some cases for something with several amperes output capability).
You are going to need high current switching regulation anyway so I wouldn't worry too much about any EMC differences. The 5A will be where the issues (if any) will tend to arise.
You will probably require a high-side switch for the 5V/5A if you can't control the regulator directly. That is most simply done with a level shifting transistor (from 3.3V to 5V) plus a power MOSFET. Something like a Diodes DMP2006UFG-7 has maximum 5.5m\$\Omega\$ Rds(on) with 4.5V drive would be okay for a high-side pass transistor. That's less than 150mW dissipation. Of course if you can use the regulator control line directly and meet your specifications it's worth avoiding extra parts. If the regulator will accept a TTL compatible input, the 3.3V may be able to control it directly (read the datasheet carefully on this point).
You could also use a MOSFET at the input of the 5V regulator which would eliminate any shutdown current it might have, but of course the micro would have to have an independent supply.
If your microcontroller draws less than 10mA or so you may find a linear regulator from the supply voltage a good solution- simple and reliable, though some switching regulators use special techniques to get reasonable efficiency even at very low output currents. Again, a trade-off to be made. If your micro draws uA in a sleep mode most of the time, a switching regulator is not going to be of any help, and when the 25W LEDs turn on a linear regulator won't make much difference even if the micro draws tens of mA.
Best Answer
Yes, you are correct, if you have the correct polarity of your multimeter leads you should at a minimum see a diode drop between Pin 8 to Pin 1 of PQ301, best case a low ohm short.