I'll use $\mathrm X^*$ instead of $X$ for character lattices, since I can never remember which is which in the $X$/$Y$ notation. I have also updated this answer from its original wrong formulation to a hopefully correct one.

$\DeclareMathOperator\srank{srank}$Note that $\Lambda_{G, P}$ is a lattice of rank $\srank(G) - \srank(M)$, where $\srank$ stands for the semisimple rank.

Exactly as written, the answer is 'no'; for example, if $G = M$ is a non-trivial torus, then $\Lambda_{G, P}$ is trivial but $\mathrm X^*(\mathrm Z(M)^\circ) = \mathrm X^*(G)$ is not.

If $G$ is semisimple, then $\srank(G) - \srank(M) = \dim(\mathrm Z(M)^\circ)$, so that $\Lambda_{G, P}$ and $\mathrm X^*(\mathrm Z(M)^\circ)$ are lattices of the same rank, hence abstractly isomorphic. However, there is a natural map $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M)^\circ)$ given by restriction, and it need not be an isomorphism; its image may have finite index. Consider the case $G = \mathrm{SL}_2$ and $M = T$.

$\DeclareMathOperator\Span{Span}$If $G$ is adjoint, then $\Lambda_G = \mathrm X^*(T)$ and $\Span_{\mathbb Z} \{\alpha : \alpha \in I(M, B \cap M, T)\}$ is the annihilator of $\mathrm Z(M)$ in $\mathrm X^*(T)$, so that the restriction map $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M))$ is an isomorphism. Since $\Lambda_{G, P}$ is torsion free, so is $\mathrm X^*(\mathrm Z(M))$, which means that $\mathrm Z(M)$ is connected, and hence we have finally that $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M)) = \mathrm X^*(\mathrm Z(M)^\circ)$ is an isomorphism in this case.

theset of simple roots" after having chosen a 'Borus', surely one may also speak of "theLevi subgroup" after having chosen a torus? @S.D. must have meant that $P$ contains $T$.) $\endgroup$