BCW71 NPN transistor as a switch
Yup.  The only way for there to be more than about 0.7V between the base and the emitter is for the transistor to be broken.
Put another way, the emitter is always about 0.7V below the base when current flows through the emitter.  The base is at 3.3V, so the emitter will be at about 2.6V.
Since you are driving a MOSFET gate with this, you need to rearrange things slightly.
Like this:

\$Y\$ goes to your 3.3V signal.  \$\overline{Y}\$ goes to R23 and then to the gate of your MOSFET.
This circuit inverts the driving signal.  A high on the base will put a low on the gate of the MOSFET.
You can either invert the logic driving \$Y\$ in code, or add another NPN in front of this one to invert the signal.