I'll use $\mathrm X^*$ instead of $X$ for character lattices, since I can never remember which is which in the $X$/$Y$ notation.

Of course you mean: is the natural map $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M)^\circ)$ an isomorphism? As you point out, my answer below is to a wrong version of this question in which the numerator of your $\Lambda_{G, P}$ is replaced by $\mathrm X^*(T)$. I will think more and update the answer soon.

The answer is 'no'; the kernel is isomorphic to $\mathrm X^*(\mathrm Z(M)/\mathrm Z(M)^\circ)$. Probably a better way to answer is that the 'correct' map is $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M))$. Consider, for example, the case where $M = G = \mathrm{SL}_2$, where $\mathrm Z(M)^\circ$ is trivial but $\Lambda_{G, P}$ is isomorphic to $\mathrm X^*(\mathrm Z(M)) \cong \mathrm X^*(\mu_2) \cong \mathbb Z/2\mathbb Z$.

In general, if $D$ is a subgroup of $T$, then $\mathrm X^*(T)/\operatorname{Ann}(D) \to \mathrm X^*(D)$ is an isomorphism, where $\operatorname{Ann}(D)$ is the space of characters trivial on $D$. The space of characters annihilating $\mathrm Z(M)$ is $\operatorname{Span}_{\mathbb Z} \{\alpha : \alpha \in \operatorname{Roots}(M, T)\} = \operatorname{Span}_{\mathbb Z} \{\alpha : \alpha \in I(M, B, T)\}$.

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.

Of course you mean: is the natural map $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M)^\circ)$ an isomorphism? The answer is 'no'; the kernel is isomorphic to $\mathrm X^*(\mathrm Z(M)/\mathrm Z(M)^\circ)$. Probably a better way to answer is that the 'correct' map is $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M))$. Consider, for example, the case where $M = G = \mathrm{SL}_2$, where $\mathrm Z(M)^\circ$ is trivial but $\Lambda_{G, P}$ is isomorphic to $\mathrm X^*(\mathrm Z(M)) \cong \mathrm X^*(\mu_2) \cong \mathbb Z/2\mathbb Z$.

In general, if $D$ is a subgroup of $T$, then $\mathrm X^*(T)/\operatorname{Ann}(D) \to \mathrm X^*(D)$ is an isomorphism, where $\operatorname{Ann}(D)$ is the space of characters trivial on $D$. The space of characters annihilating $\mathrm Z(M)$ is $\operatorname{Span}_{\mathbb Z} \{\alpha : \alpha \in \operatorname{Roots}(M, T)\} = \operatorname{Span}_{\mathbb Z} \{\alpha : \alpha \in I(M, B, T)\}$.

I'll use $\mathrm X^*$ instead of $X$ for character lattices, since I can never remember which is which in the $X$/$Y$ notation.

Of course you mean: is the natural map $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M)^\circ)$ an isomorphism? As you point out, my answer below is to a wrong version of this question in which the numerator of your $\Lambda_{G, P}$ is replaced by $\mathrm X^*(T)$. I will think more and update the answer soon.

The answer is 'no'; the kernel is isomorphic to $\mathrm X^*(\mathrm Z(M)/\mathrm Z(M)^\circ)$. Probably a better way to answer is that the 'correct' map is $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M))$. Consider, for example, the case where $M = G = \mathrm{SL}_2$, where $\mathrm Z(M)^\circ$ is trivial but $\Lambda_{G, P}$ is isomorphic to $\mathrm X^*(\mathrm Z(M)) \cong \mathrm X^*(\mu_2) \cong \mathbb Z/2\mathbb Z$.

In general, if $D$ is a subgroup of $T$, then $\mathrm X^*(T)/\operatorname{Ann}(D) \to \mathrm X^*(D)$ is an isomorphism, where $\operatorname{Ann}(D)$ is the space of characters trivial on $D$. The space of characters annihilating $\mathrm Z(M)$ is $\operatorname{Span}_{\mathbb Z} \{\alpha : \alpha \in \operatorname{Roots}(M, T)\} = \operatorname{Span}_{\mathbb Z} \{\alpha : \alpha \in I(M, B, T)\}$.