An involutive local frame can be reduced to one field plus commuting transverse fields

By the flow-box theorem there are local coordinates [given] $(t,u^2,\dots ,u^n)$ centered at $p$ in which $X_1={\mathrm{\partial }}_t$. Shrinking if necessary, write $X_j=a_j{\mathrm{\partial }}_t+{\sum }_{m=2}^n{b}_{jm}{\mathrm{\partial }}_{u^m}(2\le j\le k)$ and define $Z_j:=X_j-a_j{X}_1={\sum }_{m=2}^n{b}_{jm}{\mathrm{\partial }}_{u^m}\mathrm{.}$ Then each $Z_j$ is tangent to $\mathcal{D}$, has no ${\mathrm{\partial }}_t$-component, and $X_1,Z_2,\dots ,Z_k$ still form a local frame of $\mathcal{D}$.

Because $\mathcal{D}$ is involutive, each bracket $[X_1,Z_j]$ is tangent to [given] $\mathcal{D}$. It also has no ${\mathrm{\partial }}_t$-component, since $X_1={\mathrm{\partial }}_t$ and $Z_j$ has none. Therefore there are smooth functions $c_{j\mathrm{\ell }}$ such that $[X_1,Z_j]={\sum }_{\mathrm{\ell }=2}^k{c}_{j\mathrm{\ell }}{Z}_{\mathrm{\ell }}(2\le j\le k)\mathrm{.}$ For each fixed transverse coordinate, solve the matrix ODE ${\mathrm{\partial }}_{t}B=-C^{T}B,B(0,u)=I_{k-1},$ where $C=(c_{j\mathrm{\ell }})$. After shrinking again, the solution matrix $B$ is smooth and invertible.

For $j=2,\dots ,k$, set $Y_j:={\sum }_{\mathrm{\ell }=2}^k{B}_{\mathrm{\ell }j}{Z}_{\mathrm{\ell }}$. Because [given] the $Y_j$ are invertible linear combinations of $Z_2,\dots ,Z_k$, the fields $X_1,Y_2,\dots ,Y_k$ form a local frame of $\mathcal{D}$. Using the Leibniz rule for brackets with function coefficients and the differential equation for $B$, one gets $[X_1,Y_j]={\sum }_{\mathrm{\ell }=2}^k({\mathrm{\partial }}_t{B}_{\mathrm{\ell }j})Z_{\mathrm{\ell }}+{\sum }_{\mathrm{\ell }=2}^k{B}_{\mathrm{\ell }j}[X_1,Z_{\mathrm{\ell }}]=0.$ Hence each $Y_j$ commutes with $X_1$.

Therefore, after shrinking the neighborhood, there is a local frame [given] $X_1,Y_2,\dots ,Y_k$ of $\mathcal{D}$ with $[X_1,Y_j]=0$ for all $j\ge 2$.