The difference between MSK as CPFSK and OQPSK
The answer to your question is yes, these two representations are equivalent. The first important thing to note is that in your formula, $d_k$ must be either $1$ or $-1$, because the frequency difference for MSK must be $\Delta f=1/2T$.

I'll add some information showing the equivalence of the two representations. Expanding your first formula gives

$$\begin{align}s(t)&amp;=A\cos\left(2\pi f_ct+d_k\frac{\pi t}{2T}+\phi_k\right)\\&amp;=A\left[\cos(2\pi f_ct)\cos\left(d_k\frac{\pi t}{2T}+\phi_k\right)-\sin(2\pi f_ct)\sin\left(d_k\frac{\pi t}{2T}+\phi_k\right)\right]\\&amp;=A\left[\cos(2\pi f_ct)\cos\left(d_k\frac{\pi t}{2T}\right)\cos(\phi_k)-\sin(2\pi f_ct)\sin\left(d_k\frac{\pi t}{2T}\right)\cos(\phi_k)\right]\\&amp;=A\left[\cos(2\pi f_ct)\cos\left(\frac{\pi t}{2T}\right)\cos(\phi_k)-\sin(2\pi f_ct)\sin\left(\frac{\pi t}{2T}\right)d_k\cos(\phi_k)\right]\\&amp;=A\left[\cos(2\pi f_ct)\cos\left(\frac{\pi t}{2T}\right)I_k-\sin(2\pi f_ct)\sin\left(\frac{\pi t}{2T}\right)Q_k\right]\tag{1}\end{align}$$

with

$$I_k=\cos(\phi_k)\quad\text{and}\quad Q_k=d_k\cos(\phi_k)\tag{2}$$

where I've used $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$, $\sin(a+b)=\cos(a)\sin(b)+\sin(a)\cos(b)$, the fact that $\sin(\phi_k)=0$ (because $\phi_k\in\{0,\pi\})$, and the fact that $d_k\in\{1,-1\}$.

Equation $(1)$ shows that the MSK signal $s(t)$ can be represented as a QPSK signal with in-phase component $I_k$ and quadrature component $Q_k$ with sinusoidal pulse shaping. In this blog post, it is shown that $I_k$ and $Q_k$ can never change at the same time, and both of them change with a rate $1/2T$. That's why this type of modulation is called offset QPSK (OQPSK). All this is related to the phase recursion shown in Eq. $(3)$ below.

In order for the MSK signal to have a continuous phase, the phase $\phi_k$ must satisfy the following recurrence relation:

$$\phi_k=\phi_{k-1}+(d_{k-1}-d_k)\frac{k\pi}{2}\qquad(\text{mod } 2\pi)\tag{3}$$

This relation is also derived in the above mentioned post (Eq. $(7)$). Note that the formula for $\phi_k$ in your question satisfies $(3)$. However, it's probably easier to directly use the recursion $(3)$. From $(3)$ it follows that

$$\phi_k=\begin{cases}\phi_{k-1}\pm\pi,&amp;k\text{ odd and }d_k\neq d_{k-1}\\\phi_{k-1},&amp;\text{otherwise}\end{cases}\tag{4}$$

From $(4)$ it is clear that $\phi_k$ changes with a rate $1/2T$.

Finally, inverting $(2)$, i.e., expressing $d_k$ and $\phi_k$ in terms of $I_k$ and $Q_k$, is straightforward:

$$I_kQ_k=d_k\cos^2(\phi_k)=d_k\tag{5}$$

because $\cos(\phi_k)$ can only be $1$ or $-1$. Furthermore, $1-\cos(\phi_k)$ equals $0$ for $\phi_k=0$ and $2$ for $\phi_k=\pi$. Consequently,

$$\phi_k=\frac{\pi}{2}(1-I_k)\tag{6}$$

Both Equations $(5)$ and $(6)$ are correctly stated in your question.

In sum, all your formulas appear to be correct. You haven't shown your formulas for $I_k$ and $Q_k$; they should be equivalent to Eq. $(2)$. I have the suspicion that your $d_k$ are either $1$ or $0$. As mentioned above, they should be either $1$ or $-1$, otherwise $s(t)$ is no valid MSK signal. Also note that the mapping between the information bits and $d_k$ is totally irrelevant for all derivations shown above.

EDIT: In analogy with Eq. $(4)$, you can derive recursions for $I_k$ and $Q_k$ from Eqs. $(2)$ and $(3)$:

$$I_k=\begin{cases}-I_{k-1},&amp;k\text{ odd and }d_k\neq d_{k-1}\\I_{k-1},&amp;\text{otherwise}\end{cases}\tag{7}$$

$$Q_k=\begin{cases}-Q_{k-1},&amp;k\text{ even and }d_k\neq d_{k-1}\\Q_{k-1},&amp;\text{otherwise}\end{cases}\tag{8}$$

Choosing an initial phase $\phi_0=0$ results in the initial values $I_0=1$ and $Q_0=d_0$.
From $(7)$ and $(8)$ it is obvious that $I_k$ and $Q_k$ never change at the same time, hence offset QPSK.