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 $\operatorname{Span}_{\mathbb Z} \{\alpha : \alpha \in \operatorname{Roots}(M, T)\} = \operatorname{Span}_{\mathbb Z} \{\alpha : \alpha \in I(M, B, T)\}$.