Hello,
I assume the underside of the FR4 PCB is all metal? If so, this will not be a good antenna. As the current in the patch is counteracted by a same but opposite current in the image element. The image element is just 3mm below the driven element. So the produced far field per ampere current in the middle of the patch will be very low. If there were no losses, the useful bandwidth will be very low and the half wave patch resonator has very high Q factor.
By adding the lossy dielectric, all your input power will be dissipated into the dielectric. Check your radiation efficiency in the simulation (don't forget to enter dielectric loss for the FR4).
There is another thing, I assume that the 82mm size in x direction. Did you check the current distribution (use a display where you can also see the direction of current, for example a current animation)? The width is over 0.5 lambda electrically, so you may get a strange current distribution with current in the patch that produce counteracting far field.
When you need some performance, you need to increase the distance to the ground plane reduce the width so you get your main patch current in X-direction. This will yield several hundred ohms impedance when feeding at the edge.
You can feed via a quarter wave transformer or make cuts in the patch so the feedline runs somewhat into the patch to reach a point of lower impedance.
When it really must be flat, you need to remove the ground plane and make a dipole construction (use meandering or bends to reduce the physical length). Of course this will gives a radiation pattern that also goes down, but this can be acceptable. Don’t forget to add a balun function in case of a dipole like antenna.