Select power source with a mosfet and no voltage drop
The short answer is no. The way you've drawn this circuit, if VUSB is present, VIN=VBAT, and if VUSB is not present, VIN=VBAT-Vbodydiode.

I'd say you need to do research into what a FET does. The best way to think about it is a unidirectional voltage-controlled switch. That is to say, current can always pass from Source to Drain through the body diode, and if there is enough voltage difference between the Gate and the Source, then current can pass either direction through the Drain and Source.

The way many systems accomplish what you're trying to do is to put a diode on each input rail, pointing toward your device. With this, power is drawn from whichever rail is higher (the channel of other diode will not be sufficiently enhanced). The only issue with this is that it will drop your voltage rail by 0.7V or so (depending on diode selection) and if you are drawing any appreciable current, you'll be heating that diode a lot (possibly causing a failure). I don't know your application, so I can't tell you if this solution would be appropriate or not.

Edit: While doing something completely unrelated, I happened across this schematic:
http://www.synthrotek.com/wp-content/uploads/2014/08/DIRT_Schematic_2.2.pdf
The two 1N418's in the bottom left corner are doing exactly what you're talking about. The 9V battery is always lower than the 12V coming from the EURO connector (note that this voltage is external knowledge, I am not able to tell that from this schematic alone), thus if the battery is the only power source present, it is used, but if the EURO power is there, then that will be used. Also note that the expected power dissipation in this device is quite low.