Proving that there is no scheme which would be perfectly CPA-secure A private-key encryption scheme Π = (Gen, Enc, Dec) has perfectly indistinguishable encryptions under a chosen-plaintext attack, if for all probabilistic polynomial-time adversaries A it holds that
Pr[PrivK_cpa(n) = 1] = 1/2. I need to show that there is no encryption scheme that can satisfy this definition.

The perfect indistinguishably under a chosen-plaintext attack: Pr[PrivK_cpa = 1] = 1/2

implicates

the perfect indistinguishably in the presence of eavesdropper: Pr[PrivK_eav = 1] = 1/2.

Which, in turn, implicates perfect secrecy: Pr[Enc_K (m) = c] = Pr[Enc_K (m') = c] for all m, m' from M, c from C.

Is my argumentation correct to this point? If so, could you, please, tell how can I now find the contradiction between the scheme being perfectly secret and perfectly indistinguishable under a chosen-plaintext attack? Should I find the contradiction using the definition of the perfect secrecy (above-mentioned) or should I use the fact that |K| should be >= |M|?

I saw this post Is there such a thing as perfect CPA security? with the same question but it does not help as there the fact that Enc is probabilistic rather than deterministic is mostly discussed.
The perfect indistinguishably under a chosen-plaintext attack: Pr[PrivK_cpa = 1] = 1/2
Is my argumentation correct to this point?

No, because the question doesn't ask about 'perfect indistinguishability' (which would include computationally unbounded adversaries), just indistinguishability against probabilistic polytime adversaries.
Assuming that the adversary has access to the public key, then here is an adversary that achieves an advantage $&gt; 1/2$:

Select a set of random coins; send them through the $Gen$ function

Compare the public key generated with the public key presented

If they are the same, use the private key generated to decrypt the ciphertext and use the resulting plaintext to make the decision.

If they are not the same, select a random bit and use that as the output.


Now, if this procedure selects a set of random coins that just happen to be the same as the coins used during the original key generation process, then this always achieves the correct answer (or, at least, with high probability if the public key system has a small probability of decryption failure); if it does not, then this gives the correct answer with probability at least 1/2 (&quot;at least&quot; because there may be other sets of random coins that result in the same public key; those would also give the correct answer).
Hence, if $\lambda$ random coins were used in the key generation process, this gives the correct answer with probability $2^{-\lambda} + 0.5(1 - 2^{-\lambda}) &gt; 1/2$