Iwasawa module and inverse limit

Yes. We first show the second claimed isomorphism. It suffices to show a canonical isomorphism $$\mathbb Z_p[\Gamma/\Gamma^{p^n}]\otimes M=M/\omega_n M.$$ Note that we have the following exact sequence $$0\to \Lambda \xrightarrow{\omega_n} \Lambda \to \mathbb Z_p[\Gamma/\Gamma^{p^n}]\to 0$$ which is seen by recalling that $\omega_n=\gamma^{p^n}-1$ generates the augmentation ideal of $\Gamma^{p^n}$. Upon tensoring this exact sequence with $M$, we obtain the needed isomorphism.

Now we need to show the first isomorphism. First, the isomorphism is clearly true when $M=\Lambda^{\alpha}$. Now choose a compact presentation of $M$: $$\Lambda^{\beta} \xrightarrow{f} \Lambda^{\alpha} \to M\to 0$$ Upon tensoring with $\Lambda/\omega_n\Lambda$ we obtain another exact sequence $$(\Lambda/\omega_n\Lambda)^\beta \xrightarrow{f} (\Lambda/\omega_n\Lambda)^\alpha \to M/\omega_nM\to 0.$$ Taking inverse limits of this, applying the case of $M=\Lambda^\alpha$, and using the fact that inverse limit is an exact functor on compact modules, we obtain $$\Lambda^\beta \xrightarrow{f} \Lambda^\alpha \to \varprojlim M/\omega_n M\to 0$$ Since this is the same presentation as before, we have $M\cong \varprojlim M/\omega_n M$.