Since $BD$ and $BF$ are tangent lines to the incircle of $\triangle ABC$, $BD = BF$. But the line $BI$ bisects $\angle DBF$, so $DF$ and $BI$ are perpendicular to each other. Similarly, $DE$ and $CI$ are perpendicular to each other. It follows that $\angle BKD = \angle BKF = 90^\circ - \angle DFK$ and $\angle CED + \angle ECI = 90^\circ$. Since $AC$ is tangent to

the excircle of $\triangle DEF$, $\angle DFK = \angle CED$, and so $\angle BKD = 90^\circ - \angle CED$. Thus $\angle BKD = \angle ECI = \angle DCI$, which means that the point $K$ lies on the excircle of the quadrilateral $CEID$. Therefore $\angle BKC = \angle IEC = 90^\circ$. In a similar way, it can be proved that $\angle BLC = 90^\circ$.
Applying Menelaus theorem to $\triangle DKL$ with respect to the line $MP$ leads to the equality $\frac{KP \cdot CJ}{JL \cdot MK} = 1$, from which we get $\frac{KP}{PC} = \frac{MK}{LM}$, as $CJ = JL$. Since $DI$ and $DM$ bisect the internal angle and external angle of $\triangle DKL$ at $D$, respectively, $\frac{KQ}{QL} = \frac{KD}{DL} = \frac{KM}{ML}$ holds. Combining this with $\frac{KP}{PC} = \frac{MK}{LM}$, we get $\frac{KP}{PC} = \frac{KQ}{QL}$. It follows that $PQ$ and $CL$ are parallel to each other.
Let $A'$ be the intersection point of $BL$ and $CK$. Note that $I$ is the orthocenter of the $\triangle A'BC$. It can easily be seen that $\angle BA'D = \frac{1}{2}\angle BCA$ and $\angle CA'D = \frac{1}{2}\angle ABC$, and so $\angle BA'C = \frac{1}{2}(\angle ABC + \angle ACB)$. It follows that $\angle KPQ = \angle A'CL = 90^\circ - \angle BA'C = \frac{1}{2}\angle BAC = \angle IAB$. Since the points $A'$, $L$, $C$, $D$ are concyclic, $\angle A'KL = \angle A'BC$, from which we obtain $\angle PKQ = 90^\circ + \frac{1}{2}\angle ACB = \angle AIB$. Since $\angle KPQ = \angle IAB$ and $\angle PKQ = \angle AIB$, $\triangle KPQ$ and $\triangle IAB$ are similar to each other, and so
$$\frac{PQ}{QK} = \frac{AB}{BI'}$$
from which the conclusion follows.