$W_0^{1,p}(\Omega)\neq W^{1,p}(\Omega)$
Since $\partial \Omega$ is smooth, $u \in W^{1,p}_0(\Omega)$ if and only if $u \in W^{1,p}(\Omega)$ and $Tu=0$ on $\partial \Omega$ where $T$ is the trace operator. If $u \in W^{1,p}(\Omega) \cap C(\overline{\Omega})$ then $Tu= u \vert_{\partial \Omega}$. Thus, any function $u \in W^{1,p}(\Omega) \cap C(\overline{\Omega})$ that is not identically zero on $\partial \Omega$ is in $W^{1,p}(\Omega)$ but $W^{1,p}_0(\Omega)$.
For example, take $u\equiv 1$ in $\Omega$.