My answer is taken from the following paper http://cseweb.ucsd.edu/~mihir/papers/oem.pdf (it is a well-written paper, so if you are interested in this topic I highly recommend you to read it).
Suppose $\mathcal{SE = (K, E, D)}$ is INT-PTXT secure. From $\mathcal{SE}$ we create another scheme $\mathcal{SE}_2 = (\mathcal{K}, \mathcal{E}_2, \mathcal{D}_2)$ which is also INT-PTXT secure, but not INT-CTXT secure. The idea is to add to every ciphertext produced by $\mathcal{E}$ a redundant bit which will be ignored upon decryption. This allows us to easily create new ciphertexts from old ones which still decrypts to the same plaintext. Intuitively, an attacker cannot exploit this against INT-PTXT (which requires it to submit ciphertexts that decrypts to new plaintexts), but it will help us to break INT-CTXT security.
In detail, the definitions of $\mathcal{E}_2$ and $\mathcal{D}_2$ are as follows.
$\underline{\mathcal{E}_2(K, M)}:\\\quad C \gets \mathcal{E}(K, M)\\\quad \text{Return } 0 || C$
$\underline{\mathcal{D}_2(K, C)}:\\\quad\text{Parse $C$ as $b || C'$ where $b$ is a bit (return $\perp$ if the parsing fails)}\\ \quad M \gets \mathcal{D}(K,C') \\ \quad \text{Return } M $
I leave it to you to prove that $\mathcal{SE}_2$ is also INT-PTXT (the reduction can be found in the last paragraph before Section 4 in http://cseweb.ucsd.edu/~mihir/papers/oem.pdf).
Here is a simple attack that shows that $\mathcal{SE}_2$ is not INT-CTXT.
$\text{Adversary $\mathcal{A}$}\\ \quad \text{Let $M$ be any arbitrary plaintext} \\ \quad C \gets \mathbf{EncryptOracle}(M)\\ \quad \text{Parse $C$ as $0 || C'$}\\ \quad \mathbf{VerificationOracle}(1 || C')$
Note that $\mathcal{A}$ is a valid INT-CTXT adversary since it asks for the verification of a ciphertext it did not previously obtain from an encrypt query.
But it is not a valid INT-PTXT adversary since both $0 || C'$ and $1 || C'$ decrypt to the same plaintext $M$ (and $\mathcal{A}$ previously asked for the encryption of $M$).
