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
$$\begin{align}s(t)&=A\cos\left(2\pi f_ct+d_k\frac{\pi t}{2T}+\phi_k\right)\\&=A\left[\cos(2\pi f_ct)\cos\left(d_k\frac{\pi t}{2T}+\phi_k\right)-\sin(2\pi f_ct)\sin\left(d_k\frac{\pi t}{2T}+\phi_k\right)\right]\\&=A\left[\cos(2\pi f_ct)\cos\left(d_k\frac{\pi t}{2T}\right)\cos(\phi_k)-\sin(2\pi f_ct)\sin\left(d_k\frac{\pi t}{2T}\right)\cos(\phi_k)\right]\\&=A\left[\cos(2\pi f_ct)\cos\left(\frac{\pi t}{2T}\right)\cos(\phi_k)-\sin(2\pi f_ct)\sin\left(\frac{\pi t}{2T}\right)d_k\cos(\phi_k)\right]\\&=A\left[\cos(2\pi f_ct)\cos\left(\frac{\pi t}{2T}\right)I_k-\sin(2\pi f_ct)\sin\left(\frac{\pi t}{2T}\right)Q_k\right]\tag{1}\end{align}$$
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
$$\phi_k=\begin{cases}\phi_{k-1}\pm\pi,&k\text{ odd and }d_k\neq d_{k-1}\\\phi_{k-1},&\text{otherwise}\end{cases}\tag{4}$$
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 $(3)$:
$$I_k=\begin{cases}-I_{k-1},&k\text{ odd and }d_k\neq d_{k-1}\\I_{k-1},&\text{otherwise}\end{cases}\tag{7}$$
$$Q_k=\begin{cases}-Q_{k-1},&k\text{ even and }d_k\neq d_{k-1}\\Q_{k-1},&\text{otherwise}\end{cases}\tag{8}$$
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.
