Yes. If
g^a mod p = g^x mod p
Then
g^ab mod p = g^xb mod p
In particular
(g^b mod p)^a mod p= (g^b mod p)^x mod p
So you can, for all practical purposes, pretend that Alice's private key is x. This is why it is important for g to be a generator of the group, so that there are no such 'sibling' private keys.
She can't know the 'original' a, but she does know that it's in the (usually) small set of numbers that differ from x by multiples of o(g). Which one it is doesn't really matter.
In this particular case, what's happening is that g
is of order 296
instead of 592
. Because of this, the actual secret key Alice chose - 530
, has a 'sibling':
c = 530 + 296 mod 592 = 234