Your pulse supply is 24V. The voltage drop on the LEDs is typically 3 to 3.5 V. Therefore, there will be 12 to 10.5V across R3, the only thing that could limit current at turn on, as the op amp and FET will be ON.
When the pulse is off, there is no current in the LEDs and R1. The op amp and the FET are driven ON.
When the pulse turns on, there is about 11V difference between the pulse supply and the LED's drop. That 11 V shows up across R1. The opamp and FET looks like a switch, because the 11 V on R3 overdrives the inputs.
Now that the opamp sees 11V at the - input, it turns off, again, overdriven
I suggest you determine what current you want through the LEDs for the luminosity you want.
Then insert a resistor where R = 11V / desired current between the pulse source V2 and the first LED D1.
I also think you should reduce the voltage of the pulse source, a the resistor I suggest you use will dissipate about 11 W at 1 A.
An alternative is to use a linear current limit circuit.
Lose the op amp, C1, R2, V3. Put a resistor, about 10k on the drain to gate of the FET.
add a small transistor, like a 2N2222A, collector to gate, base to drain and emitter to ground.
Change R1 to R1 = 0.6V/desired current limit. R1 controls the behavior of the 2N2222A. When the desired current passes through R1, the transistor starts turning on, controlling the current by controlling the gate of the FET.
When there is no current, the 2N2222A will be OFF, and the FET will ???.
When the pulse turns on, the current will rise in R1, turning the 2N2222A and the FET ON. The current in R1 will cause the transistor to pull the gate of the FET down, forcing the FET to move toward OFF, reducing the current and lowering Vbe, etc.
It regulates and cannot behave like a switch.
I assumed you chose the FET for the appropriate voltages, current , power and SOA.