Let $\phi$ be your favorite $C^\infty$ function with support in $[-1/2,1/2]$ and $\int \phi = 1$. Define $\phi^\varepsilon(x):=\varepsilon^{-1}\phi(x/\varepsilon)$ and let $\Phi^\varepsilon(x):=\sum_{k\in \mathbb{Z}} \phi^\varepsilon(x-k)$ be its periodization.
As an exercise, one can check that $\widehat{\Phi^\varepsilon}(t)=\widehat{\phi}(\varepsilon t)$. Now rescale $\Phi^\varepsilon$ and define $\Phi^\varepsilon_p(x):=\Phi^\varepsilon(px)$. Expanding $\Phi^\varepsilon$ in Fourier series we have $$ \Phi^\varepsilon(x) = \sum_k \widehat{\Phi^\varepsilon}(k) e^{i2\pi kx} = \sum_k \widehat{\phi}(\varepsilon k) e^{i2\pi kx}$$ hence $$ \Phi^\varepsilon_p(x) = \sum_k \widehat{\phi}(\varepsilon k) e^{i2\pi k px} $$ which essentially shows that $\widehat{\Phi^\varepsilon_p}(k)=0$ if $p$ does not divide $k$.
I am having some troubles in getting to the same result without explicitly calling in the Fourier series (i.e. by going the "hard" way of computing $\widehat{\Phi^\varepsilon_p}$). My issue is that I end up with $\widehat{\Phi^\varepsilon_p}(x)=\widehat{\phi}(\varepsilon x/p)$, but I don't know how to conclude that $\widehat{\Phi^\varepsilon_p}(x)=0$ if $x$ is an integer and $p$ does not divide $x$. Am I missing something trivial?