Switching on DC to AC converter large voltage drop
A push-pull output such as your NE555's is ideal for driving the gate. However, your 100K gate series resistor (R3) is far too high. That resistor has to charge the gate-drain capacitance of your FQP47P06 output MOSFET.
From its datasheet, that's Ciss which is 2800 pF typ. and 3600 pF max. Working from the max. value, a very rough calculation shows it taking over 400 us to charge your MOSFET gate capacitance from the NE555 low output voltage of 0.1 V to the 2.6 V or so Vgs needed to start turning on the MOSFET.
Your 1.5 kHz output has a period of around 666 us. So the rise time and fall time of the MOSFET gate signal are much longer than the waveform period and it never sees a square wave.
Reducing the gate resistor (R3) to 470R gives, very roughly, a rise time of around 2 us. The fall time will be of a similar order, so the combined rise/fall time should be &lt;10 us, less than 2% of the gate waveform period.
The gate resistor must be large enough for the MOSFET gate current at switching to not exceed the NE555 max. output current of 200 mA. The 470R on a 12 V output will draw around  25 mA instantaneous current as a very brief spike. To ensure this current spike does not cause an NE555 supply rail dip, you should put a decoupling capacitors across its supply rail and ground, close to the NE555 +Vcc and GND pins. Use both a 10 uF and a 100 nF in parallel.