The general solution to the time-harmonic wave equation on a string may be written as $$y(x,t)=\sin(\omega t)\left[A\sin(kx) + B\cos(kx)\right].$$ The condition at $x=0$ is that $y(0,t)=A_0\sin(\omega t)$, and so we find that $B=A_0$. We further require that $y(L,t)=0$, where $L$ is the distance between $P$ and $Q$. This requirement yields $$A\sin(kL) + A_0\cos(kL) = 0,$$ or $$ A = -A_0\frac{\cos(kL)}{\sin(kL)}. $$ The total solution is then given by $$ y(x,t) = A_0\left[\frac{\sin(\omega t)\cos(kx)\sin(kL)}{\sin(kL)} - \frac{\sin(\omega t)\sin(kx)\cos(kL)}{\sin(kL)}\right]. $$ Using trigonometric identities we may then write $$y(x,t) = A_0\left[\frac{\cos(\omega t-kx+kL)}{2\sin(kL)} - \frac{\cos(\omega t +kx-kL)}{2\sin(kL)}\right].$$
Notice that the first term in the square brackets represents the forward-propagating wave (from $P$ to $Q$), while the second term represents the backward propagating wave. Further notice that the two terms have opposite sign when evaluated at $x=L$. This opposite sign could also be considered a factor of $\cos(\pi)$. Hence, the reflection at the rigid interface yields a phase factor of $\pi$.: $$y(x,t) = A_0\left[\frac{\cos(\omega t-kx+kL)}{2\sin(kL)} + \frac{\cos(\omega t +kx-kL+\pi)}{2\sin(kL)}\right].$$