Recovery of private key in AES-CBC from two ciphertexts with different IV, but identical plantext
To give a short answer, if an attacker knows ($P$, $IV_1$, $M_1$) and ($P$, $IV_2$, $M_2$) such as :

$M_1 = E_K(P, IV_1)$ 

and

$M_2 = E_K(P, IV_2)$

where $E_K$ is the AES-CBC encryption function with the key $K$ then he doesn't learn anything about $K$ because AES is resistant to known plaintext attacks. As long as the $IV$ is randomly chosen, encrypting the same message with the same key $K$ do not reduce the security of AES-CBC.

Here is a longer answer : Why is AES resistant to known-plaintext attacks?