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No, and a simple example goes as follows: $$ \begin{split} K(\varphi_{1}\otimes\cdots\otimes\varphi_{N}) & =\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\varphi_1(0)\cdot\ldots\cdot\varphi_{N}(0)\\ & \triangleq\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\delta(x_1,\ldots,x_n)\quad x_k\in\Bbb R^{n_k}, k= 1,\ldots, N \end{split} $$ where $\delta$ is the customary Dirac distribution in $\mathcal{S}'(\mathbb{R}^{n_{1}+\cdots+n_{N}})$. The distribution $K$ is then obviously a Schwartz distribution but it is not a measure, in a similar way as the example $\text{(NIF)}$ in this answer is not.