What happens if we choose the random secret keys in Diffie-Hellman greater than prime? Diffie-Hellman works as follows:

Given public parameters $p$ (a large prime) and $g$ (always referred to as a generator of $(\mathbb{Z}^∗_p)$. Then:


Alice randomly chooses $a&lt;p$ and sends $A\leftarrow g^a \mod p$ to Bob;
Bob randomly chooses $b&lt;p$ and sends $B\leftarrow g^b \mod p $ to Alice;
Alice computes $S\leftarrow B^a \mod p$;
Bob computes $S\leftarrow A^b \mod p$.


What happens if we choose $a$ and $b$ grater than $p$?
The modulus operation $\pmod p$ is performed at each step and reduces the result into $\bmod p$. And a clever implementation can use the Fermat's Little Theorem instead of taking the power than reducing to modulo $p$. After that it's possible to use the modular version of repeated squares algorithm or similar. 

Example 1) code used from  sublimerobots

sharedPrime = 23    # p
sharedBase = 5      # g

aliceSecret = 600     # a
bobSecret = 1500      # b

Alice Sends Over Public Chanel:  8
Bob Sends Over Public Chanel:  4

Privately Calculated Shared Secret:
Alice Shared Secret:  2
Bob Shared Secret:  2


Example 2)

sharedPrime = 23    # p
sharedBase = 5      # g

aliceSecret = 6000000     # a
bobSecret = 15000000   # b

Alice Sends Over Public Chanel:  8
Bob Sends Over Public Chanel:  4

Privately Calculated Shared Secret:
Alice Shared Secret:  2
Bob Shared Secret:  2


I think you are confusing the mathematical representation and the actual value. 



In the sense of optimization, the code from sublimerobots is not good. Actually. instead of 

bobSharedSecret = (A**bobSecret) % sharedPrime


a faster version

bobSharedSecret = pow(A,bobSecret,sharedPrime)


which uses modular binary exponentiation.