This is too long to be a comment.

Let $X$ and $Y$ be independent ${\rm N}(\mu,\sigma_2)$ and ${\rm N}(0,q^2)$ rv's, respectively.
Since $X+Y \sim {\rm N}(\mu,q^2+\sigma^2)$, it is equal in distribution to $Z + \mu$, where $Z \sim {\rm N}(0,q^2+\sigma^2)$. Hence,

$$
{\rm P}(X + Y \le \theta ) = {\rm P}(Z \le \theta - \mu ) = \frac{1}{{\sqrt {2\pi (\sigma ^2 + q^2 )} }}\int_{ - \infty }^{\theta - \mu } {e^{ - z^2 /[2(\sigma ^2 + q^2 )]} \,{\rm d}z} .
$$
On the other hand, by the law of total probability (conditioning on $X$), we have
$$
{\rm P}(X + Y \le \theta ) = \int_{ - \infty }^\infty {{\rm P}(Y \le \theta - x)\frac{1}{{\sqrt {2\pi \sigma ^2 } }}e^{ - (x - \mu )^2 /(2\sigma ^2 )} \,{\rm d}x}.
$$
Therefore,
$$
{\rm P}(X + Y \le \theta ) = \int_{ - \infty }^\infty {\bigg[\int_{ - \infty }^{\theta - x} {\frac{1}{{\sqrt {2\pi q^2 } }}e^{ - y^2 /(2q^2 )} \,{\rm d}y\bigg]\frac{1}{{\sqrt {2\pi \sigma ^2 } }}e^{ - (x - \mu )^2 /(2\sigma ^2 )} \,{\rm d}x} }.
$$
So,
$$
\int_{ - \infty }^\infty {\bigg[\int_{ - \infty }^{\theta - x} {\frac{1}{{\sqrt {2\pi q^2 } }}e^{ - y^2 /(2q^2 )} \,{\rm d}y\bigg]\frac{1}{{\sqrt {2\pi \sigma ^2 } }}e^{ - (x - \mu )^2 /(2\sigma ^2 )} \,{\rm d}x} },
$$
which may correspond to the left-hand side expression in the question, is equal to
$$
\frac{1}{{\sqrt {2\pi (\sigma ^2 + q^2 )} }}\int_{ - \infty }^{\theta - \mu } {e^{ - z^2 /[2(\sigma ^2 + q^2 )]} \,{\rm d}z},
$$
which may correspond to the right-hand side expression in the question (where $s^2$ should be $\sigma^2$).