Unless you are driving the FETs at a high switching frequency, or the circuit is in an application that is sensitive to very short, very small VHF noise bursts, R3 and R4 are not needed.
At first glance, it appears that R1 is much smaller than it needs to be. This leads to excessive power dissipation; 1.6 W in your case, requiring a 3 W part for good long-term reliability. Increasing it to 10 K looks fine, but it can affect the output.
Depending on the actual part numbers for the two output FETs (reference designators - !), their gate capacitances can be significant. For example, if each FET has an input capacitance of 1 nF, the time constant of waveform the gates would see is 2 us with R1 equal to 1 K, and 20 us with R1 equal to 10K. We don't know anything about the input or output waveform requirements, so only you can say if this is significant.
There is nothing in the circuit to assure balanced current sharing between the output FETs. FETs are better than bipolar transistors at self-balancing, but adding a ballast resistor in series with each source gives you more control over this. However, it also increases the circuit's output impedance, as well as the overall circuit power dissipation.
Now, for the real problem . . .
The FETs are either incorrect parts or connected incorrectly, depending on your design intent.
You show the FETs as source-followers, not saturated switches. If this is what you want, or need due to signal logic polarities, then change the FETs to n-channel types. A source-follower can act as a switch, but it is non-saturated. Thus, the voltage across them always will always be greater than their rated threshold voltage, significantly increasing their power dissipation. There is a simple circuit change that can reduce this dramatically, if you can invert the logic polarity of the input signal.
Also, as pointed out, the internal body diodes are a problem. If you reverse the source and drain connections, then the FETs act as saturated switches and the circuit needs one additional resistor (R4). R1 and R4 form a voltage divider to assure that the "gate voltage" (actually, the gate-source voltage Vgs) does not exceed the max rating for the part. For many power MOSFETs this is 20 V, with 10 V needed for maximum enhancement (complete turn-on). One way to achieve this is to leave R1 at 1 K and add in R4 at 2.2 K to 2.7 K. Again, scaling these values up by 10x reduces power dissipation and component size.
