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](https://mathoverflow.net/questions/349057/question-about-functional-derivatives/349584#349584) is not.<br>