This circuit is probably close to correct, but I can't tell without knowing what kind of load you are driving.
Most loads will cause the switching transistor to dissipate a pulse of energy during turn on. The transistor you selected can handle a pulse of 400mJ which makes it appropriate for use as a load switch. It has an on resistance of 0.3 ohms, which may be appropriate for a 1A load (although lower resistance wouldn't hurt).
Some things you should consider are...
1) How much capacitance does your load have? For example if you use this to turn on another power supply or some circuit card then most likely there will be capacitance on the front end of that setup.
a) During turn on M1 will dissipate 0.5 * C * (24V)^2 of energy. So 288uJ for every 1uF of load capacitance. Look at the safe operating area graph of the IRF9530, if the turn on time is several ms then you can dissipate up to 400mJ, which limits you to a load of 1.38mF max.
2) Your circuit doesn't control the rise rate at turn on. If your load has any capacitance whatsoever then its possible that you could exceed the 48A pulsed drain current rating.
You can fix this by adding a capacitor between the drain and gate of M1. The turn on rise rate then becomes...
dv/dt = ((24V - Vgs_th) / R2 -Vgs_th / R1) / C.
So for example C = 100nF Vgs_th=3.0V, R1=4.7K, R2=10K gives dv/dt = 14.6kv/s.
24V / (14.6kv/s) gives a rise time of 1.6ms (going from 0V to 24V).
3) The switch doesn't have any kind of short circuit protection. This may be OK if the 24V source is current limited. But if not, its worth considering.