= Solution
Because $K$ is continuous, adapted, and strictly positive, $K^{-1/2}$ is predictable and locally integrable against $M$. Define
$$
\boxed{W_t=\int_0^t K_s^{-1/2}\,dM_s,\qquad H_t=\sqrt{K_t}.}
$$
The <quadratic variation of a stochastic integral> is
$$
\langle W\rangle_t=\int_0^t K_s^{-1}\,d\langle M\rangle_s=t.
$$
Thus $W$ starts at zero and is a <continuous local martingale> with bracket $t$. The <Lévy characterization of Brownian motion> makes it a <Brownian motion> in the original filtration.
The process $H$ is continuous and adapted, and $\int_0^tH_s^2\,ds=\int_0^tK_s\,ds<\infty$ almost surely. The <associativity of stochastic integration> gives
$$
\boxed{\int_0^tH_s\,dW_s
=\int_0^t\sqrt{K_s}K_s^{-1/2}\,dM_s=M_t-M_0.}
$$
This is the <stochastic-integral representation from absolutely continuous quadratic variation>.
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