1) Let us for simplicity assume that the Hilbert space is finite-dimensional.
Let $\hat{A}:H\to H$ and $\hat{B}:H\to H$ be two Hermitian operators (a.k.a. quantum observables). Their two spectra must then be finite sets of eigenvalues
$${\rm Spec}(A)~=~\{a_1, \ldots, a_n\} \qquad {\rm and} \qquad
{\rm Spec}(B)~=~\{b_1, \ldots, b_m\}.
$$
2) In the question(v1) OP specifically asks what happens if the spectra are degenerate? We can decompose the Hilbert space
$$H~=~\oplus_{i=1}^n E^{(\hat{A})}_i$$
in orthogonal eigenspaces
$$E^{(\hat{A})}_i~:=~ {\rm ker}(\hat{A}-a_i{\bf 1}) ~\subseteq~H$$
for $\hat{A}$. Let us define corresponding projection operators $\hat{P}^{(\hat{A})}_i:H\to E^{(\hat{A})}_i$.
Then we may decompose
$$\hat{A}~=~\sum_{i=1}^n a_i\hat{P}^{(\hat{A})}_i. $$
3) Let us for simplicity assume that the initial state is a pure state given by a ket $\mid\psi\rangle$. (For the case of a mixed state, see this answer.) A measurement of the observable $\hat{A}$, with the outcome $a_i$, collapses the initial state $\mid\psi\rangle$ into a new state
$$ \mid\psi^{\prime}\rangle~=~\frac{\hat{P}^{(\hat{A})}_i\mid\psi\rangle}{\sqrt{\langle\psi\mid\hat{P}^{(\hat{A})}_i\mid\psi\rangle}}.$$
4) Similarly, we can do the same thing with the other operator,
$$H~=~\oplus_{j=1}^m E^{(\hat{B})}_j,$$
$$E^{(\hat{B})}_j~:=~ {\rm ker}(\hat{B}-b_j{\bf 1}) ~\subseteq~H,$$
$$\hat{B}~=~\sum_{j=1}^m b_j\hat{P}^{(\hat{B})}_j. $$
5) Since the two operators $\hat{A}$ and $\hat{B}$ commute $[\hat{A},\hat{B}]=0$, i.e., are compatible, there exists a common set of orthogonal eigenspaces
$$E_{ij} ~:=~ E^{(\hat{A})}_i \cap E^{(\hat{B})}_j~\subseteq~H, \qquad\qquad H~=~\oplus_{i=1}^n\oplus_{j=1}^m E_{ij}.$$
Note that some of the subspaces $E_{ij}$ could be trivial (zero-dimensional). All the projection operators $\hat{P}^{(\hat{A})}_i$ and $\hat{P}^{(\hat{B})}_j$ will also commute.
6) If we next perform a measurement of the observable $\hat{B}$, with result $b_j$, on the state $\mid\psi^{\prime}\rangle$ from section 3, the state collapses into
$$ \mid\psi^{\prime\prime}\rangle
~=~\frac{\hat{P}^{(\hat{B})}_j\mid\psi^{\prime}\rangle}{\sqrt{\langle\psi^{\prime}\mid\hat{P}^{(\hat{B})}_j\mid\psi^{\prime}\rangle}},$$
which after some straightforward algebra reduces to
$$ \mid\psi^{\prime\prime}\rangle
~=~\frac{\hat{P}^{(\hat{B})}_j\hat{P}^{(\hat{A})}_i\mid\psi\rangle}{\sqrt{\langle\psi\mid\hat{P}^{(\hat{B})}_j\hat{P}^{(\hat{A})}_i\mid\psi\rangle}}. $$
7) The latter expression is symmetric in $\hat{A} \leftrightarrow \hat{B}$,
and therefore performing the measurements in opposite order, i.e., measuring first $\hat{B}$, with outcome $b_j$, and then $\hat{A}$, with result $a_i$ would produce the same final state $\mid\psi^{\prime\prime}\rangle.$
8) So to answer the question, in the degenerate case there may still be a collapse associated with the second measurement. However, if the spectrum of the first observable is non-degenerated, then there is no second collapse.
