In CBC mode if 2 cipher blocks are same then is CBC still secure? I was going through this question Why is it said that if we have a duplicate ciphertext block it can leak our information? and I was just wondering


if there is a $P_i \oplus P_j= C_{i−1}\oplus C_{j−1}$ relationship, is CBC still secure?
How does this equality lead to information leak?
if there is a $P_i \oplus P_j= C_{i−1}\oplus C_{j−1}$ relationship, is CBC still secure?
  


As commented by SqueamishOssifrage standard collision of $2^{64}$ blocks for a 128-bit block cipher is not a hard limit. We want a negligible success for the adversaries. With $q$ queries ( encryptions) the advantage of the adversary must be negligible like $2^{-32}$ or $2^{-64}$.

To achieve this we need 


$q^2/2^{128} = 1/2^{64}$ that is $q=2^{32}$, or
$q^2/2^{128} = 1/2^{32}$ that is $q=2^{48}$


Therefore you should not encrypt more than $2^{32}$ or $2^{48}$ block depending on the advantage settings of the adversary.

The CBC mode has more problems than this;


The IV must be unpredictable.
The padding oracle attacks on servers.
The BEAST and Lucky13 attacks on TLS only.


And, therefore CBC mode is removed from TLS 1.3. CBC mode is archaic. Today, you should use Authenticated Encryption as AES-GCM or ChaCha20-Poly1305.


  
  How does this equality lead to information leak?
  


You have a relation $P_{i}\oplus P_j=C_{i-1}\oplus C_{j-1}$ in which you know 
$C_{i-1} \text{ and } C_{j-1}$ then you know the $\oplus$ difference of the two plaintext. In AES, you will have at most 128-bit to apply crib-dragging attack like in OPT. If you know one of the plaintext you know the other.