Pauli-forbidden term symbols for atomic carbon
Okay, actually I managed to find somewhat of an explanation myself.

We could construct a state with $M_L = 1$ and $M_S = 0$, for example:
$$\begin{align}
\psi = \frac{1}{2}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2) + \mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right]
\end{align}$$

where $\mathrm{2p}_m$ represents the 2p orbital with $m_l = m$.

However, such a combination of $(M_L,M_S)$ can only occur once, since the two electrons are indistinguishable. In other words, the following two other wavefunctions are not permissible because permuting the two electrons changes the wavefunction:
$$\begin{align}
\psi &amp;= \frac{1}{\sqrt{2}}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right] \\
\psi &amp;= \frac{1}{\sqrt{2}}\left[\mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right]
\end{align}$$

and the $^1D$ state, which has $S = 0 \Rightarrow M_S = 0$ and $L = 2 \Rightarrow M_L = 2,1,0,-1,-2$, already has to encompass one state with such a combination of $(M_L,M_S)$. Therefore, a state with $(M_L,M_S) = (1,0)$ cannot contribute to a $^1P$ term.

It is logically not the strongest argument (sufficiency vs necessity), but it is kind of what I was looking for. If anybody can provide another viewpoint though I will be glad to accept the answer.