= Solution
Assume $K\geq1$ is an integer, $0<p<1$, and put $q=1-p$. There is a discrepancy between the printed state definition and the formula requested later. We first follow the stated cumulative count, which retains previous good items when a defective item appears during phase A. The transition probabilities are
$$
P_{i,i}=p,\quad P_{i,i+1}=q\quad(0\leq i<K),\qquad
P_{K,0}=p/r,\quad P_{K,K}=1-p/r,
$$
with all others zero. The balance equations imply $q\pi_i=q\pi_{i-1}$ for $1\leq i<K$ and $(p/r)\pi_K=q\pi_{K-1}$. Thus
$$
\boxed{\pi_i=\frac p{Kp+rq}\quad(0\leq i<K),\qquad\pi_K=\frac{rq}{Kp+rq}}.
$$
For comparison, if clearance requires $K$ consecutive good items, each defect in phase A resets the state to zero. Replace its transitions by $P_{i,0}=p$, $P_{i,i+1}=q$, combining contributions if destinations coincide. Phase B retains its two transitions. For this <consecutive-clearance inspection model>, balance yields $\pi_i=q^i\pi_0$ and $(p/r)\pi_K=q^K\pi_0$. With $D=1+(r-1)q^K$, normalization gives
$$
\boxed{\pi_i=\frac{pq^i}{D}\quad(0\leq i<K),\qquad\pi_K=\frac{rq^K}{D}}.
$$
This second model is the one giving the printed formula in part (iii). It is a different transition rule, rather than an algebraic simplification of the cumulative model.
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