Saudi Booklet

Let $O$ be the circumcentre of triangle $ABC$. First we note from standard properties of the Miquel point $S$, we have:
$$\bullet \ \triangle SM_C M_B \sim \triangle SBC \sim \triangle SPQ; (*)$$
* I and S are inverses with respect to circle $ABC$;
* $\angle OSX = 90^\circ$.

From the above we have $\triangle OM_A I \sim \triangle OSM_A$ and
$$\angle M_A PB = \angle M_A SX = 90^\circ + \angle M_A SO = 90^\circ + \angle OM_A I = \angle M_A BA = \angle CDA.$$
Observe that $\angle PM_C M_A = \angle BM_C M_A = \angle DAC$ and $\angle M_C M_A B = \angle ICD$. Combining these with $\angle M_A PB = \angle CDA$ shows $M_C PM_A B \sim ADCI$. Therefore, we have $\frac{M_C B}{BP} = \frac{AI}{ID}$. Similarly, $\frac{M_B C}{CQ} = \frac{AI}{ID}$. Thus we get
$$\frac{M_C B}{BP} = \frac{M_B C}{CQ} = \frac{AI}{ID}. \qquad (\text{**})$$
Now, observe that $\angle ICB = \angle AM_B M_C$ and $\angle CBI = \angle M_B M_C A$ which gives that $\triangle IBC \sim \triangle AM_C M_B$. This, combined with (**), is enough to show $\triangle DPQ \sim \triangle IBC$ by linearity, thus $\frac{DP}{DQ} = \frac{IB}{IC}$.
Combining $\triangle IBM_C \sim \triangle ICM_B$ with (***) shows $IBM_C P \sim ICM_B Q$, thus $\frac{IP}{IQ} = \frac{IB}{IC}$. Finally, we have that
$$\frac{SP}{SQ} = \frac{SB}{SC} = \frac{BM_C}{CM_B} = \frac{IB}{IC}$$
from (*) and $\triangle IBM_C \sim \triangle ICM_B$. Putting this together with above ratios, we have
$$\frac{IB}{IC} = \frac{DP}{DQ} = \frac{IP}{IQ} = \frac{SP}{SQ}$$
which shows that circle $SID$ is an Apollonius circle with respect to $P$ and $Q$, giving the desired conclusion.