Recall that the adversary in the IND-CPA security game gets a "left-right oracle":
$$\mathsf{LR}(m_0,m_1) := \mathsf{Enc}_{pk}(m_b)$$
where $b\in\{0,1\}$ is a secret bit that the adversary has to try to recover.
suppose that there is an adversary who can find that the two ciphertexts are encrypted from the same plaintext.
In the above, you can see that there is only one ciphertext though --- $\mathsf{LR}(m_0,m_1)$ returns the single ciphertext $\mathsf{Enc}_{pk}(m_b)$. This means I am not 100% sure of what you intend to ask. If we modify your question to:
Suppose there is an adversary $\mathcal{A}$ who, given two arbitrary ciphertexts $c, c'$ encrypted under the same public key, can determine whether they encrypt the same thing --- meaning can answer the question
$$\mathsf{Dec}_{sk}(c)\stackrel{?}{=}\mathsf{Dec}_{sk}(c')$$
Can this adversary break IND-CPA security?
The answer is yes.
The attack is simple --- for distinct $m_0, m_1, m_2$, query $c\gets \mathsf{LR}(m_0, m_1)$, $c'\gets \mathsf{LR}(m_2, m_1)$. If the adversary determines that $c, c'$ encrypt the same message, then both $\mathsf{LR}$ queries encrypted $m_1$, e.g. $b = 1$. Otherwise, $b = 0$.
This is actually the foundation of a fairly fundamental result in public-key encryption.
Deterministic public-key encryption is impossible.
I think the citation for this is Goldwasser-Micali 1982, but regardless of the particulars it is one of the "early" results in theoretical cryptography.
To prove this, all you need to do is notice that if $\mathsf{LR}$ is deterministic, building the adversary $\mathcal{A}$ we mentioned before is quite simple (I'll let you think of how).
We can then mount the attack we mentioned before.
