No, it is not a misprint. By definition,
$$\mu(E) := p_t(x,E)$$
defines a measure for each fixed $x$ and $t \geq 0$. The integral
$$\int_K p_t(x,dy) p_s(y,E)$$
is the integral of the mapping $y \mapsto p_s(y,E)$ with respect to the measure $\mu$, i.e.
$$\int_K p_t(x,dy) p_s(y,E) = \int_K p_s(y,E) \mu(dy).$$