A large Schottky diode by itself will give you about a half volt drop in the forward direction. (What is the purpose of the transistor circuit after the diode)?
If you want less drop then that you could use a power MOSFET. A simulation of such a circuit is shown below.
Since a MOSFET conducts equally well in both directions, the P-MOSFET is shown connected with the input going to the drain so that it can block the reverse voltage when the drain becomes negative. For normal operation, when the drain goes positive, the MOSFET substrate diode initially conducts current and the output voltage starts to rise. When the voltage is high enough that the gate-source voltage exceeds the MOSFET threshold, it starts to turn on. With 24V applied, R1 and R3 reduce the gate voltage to 12V so it won't exceed the maximum Vgs rating of the MOSFET.
When the voltage is reversed, the MOSFET substrate diode is reversed biased and the MOSFET stays biased off (Vgs = 0V) thus the MOSFET blocks the reverse voltage as desired.
The MOSFET shown has an ON resistance of 3.6mΩ, giving a voltage drop of about 55mV @ 16A as shown by Cursor 1 in the simulation. You can use a different P-MOSFET as long as the current and voltage rating are greater than the input voltage, and the ON resistance gives you the desired maximum voltage drop when conducting the load current.
Note that if you can connect the circuit in the ground (negative) lead of the circuit being protected then you can use an N-MOSFET which, due to the higher N material mobility, allows the use of a smaller chip and cheaper device for the same current and ON-resistance rating.