Let $H$ be any algebraic group (I don't know what framework you are using, but affine and finite type, and smooth (although this is automatic over $\mathbb{C}$) is what I care about). Then any homomorphism $H\to \mathbb{G}_m$ sends $R_u(H)$, the unipotent radical of $H$, to a unipotent subgroup of $\mathbb{G}_m$, of which there is only the trivial group. So, in fact, you see that $M(H)=M(H/R_u(H))$.
In particular, if now $H=P$ is a parabolic with Levi factor $L$ then $M(P)=M(L)$. Note that $L$ is a reductive group, and it's easy to see that $M(L)=M(L^\mathrm{ab})$ (where $L^\mathrm{ab}$ is the abelianization of $L$). Note, moreover, that since $L$ is reductive that $Z(L)^\circ\to L^\mathrm{ab}$ (where $\circ$ denotes the connected component) is a surjection with finite kernel (an isogeny). So, if you only care about the rank of $M(L)$ then you can replace it with $M(Z(L)^\circ)$ (note that $Z(L)^\circ$ is a torus).
Example: If $P=B$ a Borel of $G$ then the above says that $M(B)=M(B/R_u(B))$ but if $T\subseteq R_u(B)$ is a maximal torus then this is (essentially) a Levi subgroup of $B$--in other words $B/R_u(B)\cong T$. Thus, $M(B)=M(T)\cong \mathbb{Z}^n$.
Example: If $P$ is the maximal parabolic of $G$, namely $P=G$, then $M(P)=M(G)=M(G^\mathrm{ab})$ and this has the same rank as $M(Z(G)^\circ)$. For example, if $G=\mathrm{GL}_n$ then $M(G)=M(G^\mathrm{ab})=M(\mathbb{G}_n)=\mathbb{Z}$ (which is the same rank as $M(Z(G)^\circ)=M(\mathbb{G}_n)=\mathbb{Z})$. But, as an extreme example, note that if $G$ is semisimple (e.g. $G=\mathrm{SL}_n,\mathrm{Sp}_{2n},...$) then $Z(G)^\circ$ is trivial and so the above shows that $M(G)=\{0\}$
Example: If $P$ is the parabolic of $\mathrm{GL}_n$ consisting of block upper diagonal matrices with sizes $\ell+k=n$ then $M(P)=M(L)$ where $L$ is the Levi factor of $P$ which is just $\mathrm{GL}_k\times\mathrm{GL}_\ell$. I leave it to you to then compute that $M(L)=M(L^\mathrm{ab})=\mathbb{Z}^2$.
