Let $\mathbb{F}$ be a finite field and $\mathbb{E}$ be an extension of a prime degree (say $q$). Let $d$ be co-prime to $|\mathbb{F}|^q-1$.

$\langle\beta_1^d,...,\beta_q^d\rangle$ is a $\mathbb{F}$-subspace of $\mathbb{E}$, whose dimension over $\mathbb{F}$ is $q$. In other words, $\{\beta_1^d,...,\beta_q^d\}$ is again a basis of $\mathbb{E}$ over $\mathbb{F}$.


Recall your question. 

At first, assume $gcd(d, p^k-1)=1$. Let $k=q_1q_2...q_m$ be a prime factorization and $k_j:=q_1q_2...q_j$.

A basis of $\mathbb{F}_{p^k}$ can be given as $\mathcal{B}_m\times \mathcal{B}_{m-1}\times ...\times \mathcal{B}_1$ where $\mathcal{B}_j$ is a basis of $\mathbb{F}_{p^{k_j}}$ over $\mathbb{F}_{p^{k_{j-1}}}$. Thus, $M_{p^k}^{(d)}=k$ in the case.

Secondly, we consider $gcd(d,p^k-1)=n\neq 1$. Then $\{\beta_1^d,...,\beta_k^d\}$ lies in a proper subspace of $\mathbb{F}_{p^k}=\{0, x~|~x^{p^k-1}=1 \pmod{p}\}$. Then we reduce your question to $d'|d$ and $k'|k$ and $gcd(d',p^{k'}-1)=1$. Here $d'=\frac{d}{n}$ and $k'$ be the maximum divisor of $k$ so that $n|\frac{p^k-1}{p^{k'}-1}$. ("maximum" or "maximal" seems to be more proper than "largest"...)


typo : Sorry for the first assumption on $d$. $q-1$ is replaced by $|\mathbb{F}|^q-1$.