Recognize whether two random values are raised to the same power

Alice selects two random numbers from a finite field $Z_p$ : $a$ and $b$.

Bob does one of the two following steps randomly (sometimes he does step 1; sometimes step 2):

1. He chooses a random number $r$ from $Z_p$ and calculates $a^r\;mod\;p$ and $b^r\;mod\;p$ and gives these two values to Alice
2. He chooses two different random numbers $r$ and $r'$ from $Z_p$ and calculates $a^r\;mod\;p$ and $b^{r'}\;mod\;p$ and gives these two values to Alice

Is there any efficient algorithm Alice can use to recognize which step Bob has done?

Little here $a$ and $b$ should be selected from $\mathbb{Z}^*_p$ and $r$ between $1$ and $p-1$ exclusive

Looking like this, it seems to relate directly to the DDH problem, at least if there exists some $w$ such that either $a = b^w$ or $b = a^w$. That is if at least one of $a$ or $b$ generates the subgroup containing the other which is a given if $p$ is a safe prime [if both are quadratic residues/non-residues either exists otherwise whichever is a non-residue is a generator], not sure about the other cases. In this case $a, b^{r'}, a^{r}$ make DDH triplet generated from $b$ or $b,a^r,b^{r'}$ make DDH triplet from $a$. In your case whether DDH problem is hard on not depends on the choice of $a$ and $b$. If both $a$ and $b$ generate the same subgroup with a large prime order then DDH problem is thought to be hard in a classical computer. So, there is no known efficient classical algorithm in that case. DDH problem can be solved with significantly higher probability than random guess in many other cases. For instance if both $a$, $b$ are non-residues modulo $p$ and among $a^r, b^{r'}$ one is a quadratic residue and another is a non residue you can tell that one of $r,r'$ is odd and another is even and thus $r \neq r'$.

In your question $a$ and $b$ are generated randomly, so I would say it is distinguishable. Not in all cases but there is advantage in the cases I mentioned earlier which in computer science is considered significant to be considered not indistinguishable

I am not sure about the case in which neither $a$ nor $b$ generates the subgroup containing the other because I cannot think of a way to relate it to DDH, at least not on top of my head. There might be some sub cases with advantage in that condition too

UPDATES: You have stated that you are trying to design a protocol. Firstly, it is not prudent to try so without a deep understanding of cryptography. Assuming that the entire security of the system relies on this, then you should use a $g$ that is used to generate $a$ and $b$ to be some generator of a large prime order subgroup $q$ and choose $r,r'$ between $1$ and $q$ to ensure that DDH problem is conjectured to be hard in classical computers. Or use known non-pairing friendly EC groups where DDH problem is conjectured to be classically hard. But still I don't know details of a protocol. And implementing it still has issues like side-channel attacks etc.

If Bob is willing to help Alice recognizing case 1, he could run a Schnorr-like protocol as a prover. He would produce a response treating $r$ as a private key exactly as specified by the protocol. Alice would verify whether this response fits both random numbers, considered as two public keys.