We will use the calculus. A non-calculus argument can be given, but it would involve drawing a number of pictures, a painful process.
Suppose that $k\ne 0$. Then a definite integral of $\sin(k(x+j))$ is $-\frac{1}{k}\cos(k(x+j))$. It follows that our integral is $0$ precisely if $\cos (k(a+j))=\cos(k(b+j))$.
There are two sorts of situations where $\cos u=\cos v$. The simplest is when $u$ and $v$ differ by a multiple of $2\pi$. So our integral will be $0$ if $k(b+j)-k(a+j)=2n\pi$ for some integer $n$. This is the case if $k(b-a)=2n\pi$, or equivalently if $b-a=n\frac{2\pi}{k}$ for some integer $n$. The congruence notation is most often used for integers. However, it can be extended, and we can write $x\equiv y\pmod{t}$ if $x-y$ is an integer multiple of $t$. Then the above result can be expressed as $b\equiv a\pmod{\frac{2\pi}{k}}$.
Note that $\cos(w)=\cos(-w)$. So we can also have $\cos u=\cos v$ if the sum of $u$ and $v$ is an integer multiple of $2\pi$. In our case, that gives the condition that $k(a+b)+2kj$ is a multiple of $2\pi$, or, in your congruence notation, $a+b+2j\equiv 0\pmod{\frac{2\pi}{k}}$. Unlike in the previous case, here the value of $j$ is relevant.