The answer is no, there is no such pseudo-Prikry sequence in the
Prikry extension.

To see this, let's first make some observations.

**Lemma 1.** If $j:V\to M$ is the ultrapower by a normal measure
$\mu$ on $\kappa$, then $$\bigcap\{\ j(C)\mid C\subset\kappa\text{
club }\}=\{\kappa\}.$$

**Proof.** Certainly $\kappa\in j(C)$ for any club
$C\subset\kappa$, and by considering final segments of $\kappa$,
it is clear that the intersection is contained in
$[\kappa,j(\kappa))$. So consider $\kappa<\alpha<j(\kappa)$. Since
$\mu$ is normal, it follows that $\alpha=j(f)(\kappa)$ for some
function $f:\kappa\to\kappa$. Let $C_f\subset\kappa$ be the club
of ordinals $\beta$ with $f''\beta\subset\beta$. It follows that
$\alpha\notin j(C_f)$, and so any particular $\alpha$ other than
$\kappa$ is cast out of that intersection. **QED**

Now consider the $\omega$-iteration of $\mu$ $$V\to M_1\to
M_2\to\cdots\to M_\omega$$

**Lemma 2.** Similarly, for the $\omega$-iteration embedding $j_\omega:V\to M_\omega$, we have $$\bigcap \{\ j_\omega(C)\mid C\subset\kappa\text{ club
in }V\ \}=\{\ \kappa_n\mid n\in\omega\ \}.$$

**Proof.** For any particular club $C\subset\kappa$, since it is
in $\mu$, we have that $\kappa_n\in j_\omega(C)$ for every $n$.
Now suppose that $\kappa_n<\alpha<\kappa_{n+1}$ for some $n$. By
standard representation results, we know that
$\alpha=j_\omega(f)(\kappa_0,\ldots,\kappa_n)$ for some function
$f:\kappa^{n+1}\to \kappa$. Let $C_f$ be the set of $\beta<\kappa$
such that $f''\beta^{n+1}\subset\beta$. This is club in $\kappa$,
but by design, $\alpha\notin j_\omega(C_f)$. So every ordinal not
on the critical sequence is cast out of the intersection. **QED**

It is a standard fact that the critical sequence
$s=\langle\kappa_n\mid n\in\omega\rangle$ is $M_\omega$-generic
for Prikry forcing with $\mu_\omega=j_\omega(\mu)$. Suppose that
$t=\langle\delta_n\mid n\in\omega\rangle$ is a pseudo-Prikry
sequence in $M_\omega[s]$. In particular, if $C\subset\kappa$ is
club in $V$, then $j(C)$ contains a tail of the $\delta_n$. But
for any particular ordinal $\alpha$ not on the critical sequence,
there is a club $C\subset\kappa$ for which $\alpha\notin
j_\omega(C)$. Thus, if infinitely many $\delta_n$ are not on the
critical sequence, we can intersect the corresponding clubs to
find a single club $C\subset\kappa$ such that $j_\omega(C)$ does
not contain a tail of $t$, which would contradict our assumption
that $t$ is a pseudo-Prikry sequence. Thus, $t$ must eventually be
contained in the critical sequence. It follows that $t$ is a
Prikry sequence over $M_\omega$, since any almost-subsequence of a
Prikry sequence is a Prikry sequence.

Thus, every pseudo-Prikry sequence in $M_\omega[s]$ is actually a
Prikry sequence, and then by elementarity, the same holds in the
Prikry forcing extension of $V$.