Can we say that if $P=NP$ there is no CPA secure public key encryption? I've learned that public key encryption is based on the problem of Discrete Log (as regard to group theory) which believed to be hard.

But, can we say that it doesn't matter on which problem our public encryption rely and say that it couldn't be CPA secure anyway?
There is no direct inference from $P = NP$ or $P \neq NP$ to security or insecurity of any particular encryption algorithm. As far as practical consequences are concerned, the "$P = NP$" problem is severely overhyped.

If $P = NP$ then any problem for which a solution can be verified in polynomial time can also be solved in polynomial time. "Polynomial time" does not automatically mean fast, for two reasons:


Complexity classes like $P$ and $NP$ are about asymptotic behaviour: they give us values for sufficiently large instances of the problem. Nothing tells us that a specific instance is large enough for the asymptotic behaviour to be an accurate description of the computational cost of solving that instance. For instance, asymptotically speaking, GNFS is a faster integer factorization algorithm than Quadratic Sieve; but, in practice, QS is faster than GNFS for integers of less than 300 bits. Proving $P = NP$ would imply that there exists a polynomial algorithm to solve integer factorization (or discrete logarithm), but it would not follow that this algorithm is indeed efficient for a 1024-bit integer.
"Polynomial" means "with cost proportional to $n^x$ for a problem of size $n$ and some constant integer $x$". Nothing guarantees that $x$ is small. A $O(n^{20})$ algorithm is completely polynomial, and completely out of reach of our technology for a size $n \geq 32$ bits...