For $p=1$ your equation indeed reduces to a simpler form. Put $X=x-t, T=t$. Then $u_t=u_T-u_X$, $u_x=u_X$, and hence in the new independent variables $X,T$ your equation (with $p=1$) becomes a well-studied linear third order equation $$ u_{T}+u_{XXX}=0, $$ which is, inter alia, a linearized version of the "standard" KdV equation with $p=2$, see e.g. the discussion in this book by Ablowitz and Segur (and cf. also Willie's answer).

For $p=1$ your equation indeed reduces to a simpler form. Put $X=x-t, T=t$. Then $u_t=u_T-u_X$, $u_x=u_X$, and hence in the new independent variables $X,T$ your equation (with $p=1$) becomes a well-studied linear third-order equation $$ u_{T}+u_{XXX}=0, $$ which is, inter alia, a linearized version of the "standard" KdV equation with $p=2$, see e.g. the discussion in this book by Ablowitz and Segur (and cf. also Willie's answer).

Note that the case of $p=3$ (the modified KdV equation) is also quite interesting. It is integrable by the inverse scattering transform just as the "usual" KdV and, in fact, is related to it through the Miura transformation (see e.g. the above book for details).