Let $K$ be a finite field extension of $\mathbb{Q}_p$. Let $G_K$ denote the absolute Galois group of $K$, $I_K$ the inertia subgroup and $I_K^{(p)}$ the $p$-Sylow subgroup of $I_K$, i.e. the wild ramification group. Let us denote the maximal unramified extension by $K^{nr}$ and the maximal tamely ramified extension by $K^{tr}$. It is true that we have $$ H^1 (G_K/I_K, \mathrm{GL}_d(\widehat{\mathbb{Q}_p^{ur}}))=0 $$ Does there exist a smiliar result for $I^{(p)}_K$  ? i.e. do we have $$ H^1 (G_K/I_K^{(p)}, \mathrm{GL}_d(\widehat{\mathbb{Q}_p^{tr}}))=0 ?$$

Let $K$ be a finite field extension of $\mathbb{Q}_p$. Let $G_K$ denote the absolute Galois group of $K$, $I_K$ the inertia subgroup and $I_K^{(p)}$ the $p$-Sylow subgroup of $I_K$, i.e. the wild ramification group. Let us denote the maximal unramified extension by $K^\text{nr}$ and the maximal tamely ramified extension by $K^{tr}$. It is true that we have $$ H^1 (G_K/I_K, \operatorname{GL}_d(\widehat{\mathbb{Q}_p^\text{nr}}))=0. $$ Does there exist a similar result for $I^{(p)}_K$? I.e. do we have $$ H^1 (G_K/I_K^{(p)}, \operatorname{GL}_d(\widehat{\mathbb{Q}_p^\text{tr}}))=0 ?$$