Thermal calculation in case of duty cycle for a mosfet
THE MOSFETs won't dissipate as much heat as you think because the series resistors shown in your circuit diagram will do that. They are intended as current limit resistors and will dissipate the main bulk of the power. However, there are significant problems in what you propose: -


LEDs directly in parallel means unequal brightness (at best) and may even mean one LED hogs most of the 1.6 amps and dies, rapidly followed by another LED taking the baton from the now dead LED and expiring itself. Pretty soon, all the LEDs are dead. 
You also have got 3 LEDs shorted out in your schematic.
The ULN2803A that you propose is only rated for 500 mA and if the current were kept to less than 500 mA it would drop about 1.5 volts and be a big source of heat.
But, you can't use the ULN2803A as a row driver because it is an NPN Darlington.
The UDN2981 is better suited for row driving (PNP Darlington) but it is also only designed for 500 mA collector current and will also produce a volt drop of around 1.5 volts (similar to the ULN2803A).


There are several issues with this proposed idea and the MOSFET power dissipation is not one of them.

However, in order to cover the bases, if you were proposing to use the IRLZ44N as a current limiter then you would need to be very careful in controlling it. For instance, if you applied a gate voltage of about 2.3 volts you would get around 1.6 amps flowing but, due to nearly instant device warming, this current would rapidly rise towards 6 amps and blow the LEDs: -



So be very careful and decide what you are trying to achieve.