Let $K$ be the foot of the perpendicular drawn from $M$ to $AC$. Let the line perpendicular to $AI$ at $I$ intersects $MK$ at $Q'$ and let $Q'I\cap BC=V$. Let $Q'C$ intersect $\Gamma$ at $T'$ for the second time. Let $\angle IAC=\angle IAB=\alpha$. We have $MN\perp BC$. We have $\angle AIQ'=\angle AKQ'=90^\circ$, hence $AIKQ'$ is cyclic. Then, $\angle IQ'K=\angle IAK=\alpha$. We have $\angle VCM=\angle BCM=\angle BAM=\alpha=\angle IQ'K=\angle VQ'M$, hence $VQ'CM$ is cyclic. We have $\angle VIM=\angle VNM=90^\circ$, so $VINM$ is cyclic. Let $\angle MIN=\angle MVN=\angle MVC=\angle MQ'C=\beta$. We have $\angle MIN+\angle IMT'=\beta+\angle AMT'=\beta+\angle ACT'=\beta+\angle KCQ'=\beta+(90^\circ-\beta)=90^\circ\Rightarrow IN\perp MT'.$ We have $IN\perp MT'$ and $T'\in \Gamma$. Hence $T'=T$. Also, $Q'$ lies on $CT$ and $\angle AIQ'=90^\circ$. Hence, $Q'=Q$. $\angle MIN=\beta=\angle MQC$ and $\angle IMN=\angle IVN=\angle QVC=\angle QMC$. Hence, $\triangle MIN\sim \triangle MQC$. Similarly, we can find that $\triangle MIN\sim \triangle MPB$. Hence, $\triangle MPB\sim \triangle MQC$. Also, we have $|MB|=|MC|$. Hence, $\triangle MPB\cong \triangle MQC\Rightarrow |PB|=|CQ|$.

My sol:- Let $P'$ be the point such that $MP'/MI$=$MB/MN$ & $\measuredangle NMB$=$\measuredangle IMP'$ & Let $Q'$ be the point such that $MQ'/MI$=$MC/MN$ & $\measuredangle NMC$=$\measuredangle IMQ'$.Note that $M$ is the center of spiral sim. sending $B-N-C$ to $P'-I-Q'$ & since $BM$=$CM$ , we get $BP'$=$CQ'$. $BP'$, $CQ'$ meet at a point ,say, $T$ on $(ABC)$.It suffices to show that $MT'$ is perp. to $NI$. Let $IN$ & $CQ'$ intersect at $D$.$D$ is foot of $M$ on $CQ'$. Let $E$ be foot of $M$ on $IN$. $\measuredangle TMD=90^\circ-\measuredangle DTM=90^\circ-\measuredangle QP'M=\measuredangle IMQ'=\measuredangle EMD$. Hence,$T'$ coincides with $T$.So, done.

Interestingly enough, is this problem generated with AI?

Let $IN \cap MT = J$, let $S$ be diametrically opposite $M$, the line through $I$ perpendicular to $AI$ intersect $BC$ at $K$ and let $KM \cap \Gamma = L$. $90 = \angle{MNK} = \angle{MIK}$ so $(KINM)$ is cyclic so $\angle{JMN} = \angle{KNI} = \angle{KMI} \Rightarrow \angle{SMT} = \angle{LMA} \Rightarrow LT || AS$ and since $AS \perp AM \Rightarrow LT || PQ$ Thus, $\angle{BPK} = \angle{BTL} = \angle{BML} = \angle{BMK} \Rightarrow (PKBM)$ is cyclic and similarly we have $(QCMK)$ cyclic. And so $\angle{KPM} = \angle{MBC}$ and $\angle{KQM} = \angle{KCM} = \angle{BCM} \Rightarrow \angle{QPM} = \angle{PQM}$ so $IP = IQ$ so if $PQ \cap AB = A'$ and $PQ \cap AC = B'$ then $PB' = QC'$. Moreover, $\angle{QC'C} = \angle{PB'B} \Rightarrow \sin\angle{QC'C} = \sin\angle{PB'B}$ and $\angle{QCC'} = \angle{TCA} = \angle{TBA} = 180 - \angle{PBB'} \Rightarrow \sin\angle{QCC'} = \sin\angle{PBB'}$. By Law of Sines, $QC = \frac{C'Q\sin\angle{CC'Q}}{\sin\angle{QCC'}}$ and $PB = \frac{B'P\sin\angle{BB'P}}{\sin\angle{PBB'}} \Rightarrow QC = PB$ as desired.

Claim: $\triangle MBP$ and $\triangle MCQ$ are similar. Proof. Let $E=IN\cap BT$. Now $\angle NMB=90^\circ-\angle A/2=90^\circ-\angle MTB=\angle TEN$, which means $MNEB$ is cyclic. So $90^\circ=\angle MNB=\angle MEB=\angle MEB$. Therefore $PIEM$ is cyclic. So $\triangle MPB\sim\triangle MIN$ by spiral similarity. Let $F=IN\cap CT$. Now $\angle NMC=90^\circ-\angle A/2=90^\circ-\angle CTM=\angle CFN$, which menas $CNMF$ is cyclic. So $90^\circ=\angle MFC=\angle MNC=\angle MIW$. Therefore $IMFQ$ is cyclic. So $\triangle MIN\sim\triangle MQC$ by spiral similarity.$\blacksquare$ Since $MB=MC$ so they are congruent. This implies $PB=CQ$.

Let $MN$ meet $\Gamma$ again at $L$ and let $LI$ meet $\Gamma$ at $T_A$. Finally, let $PQ$ meet $AB$ and $AC$ at $B_1$ and $C_1$ respectively. Note that $\angle MIN = \angle ILM$. Thus $\angle LMT_A = \angle AMT$, thus $T_AL = AT$. Angle chasing yields \[ \angle BPB_1 = \angle BAT_A \qquad \text{and} \qquad \angle C_1QC = \angle T_AAC. \]Thus, by law of sines we have \[ \frac{PB}{BB_1} \cdot \frac{CC_1}{CQ} = \frac{\sin \angle C_1B_1A}{\sin \angle BPB_1} \cdot \frac{\sin \angle C_1QC}{\sin \angle AC_1B_1} = \frac{\sin \angle C_1QC}{\sin \angle BPB_1} = \frac{T_AC}{T_AB}. \]However, note that $T_A$ is the point were the $A$-mixtilinear incircle of $ABC$ touches $(ABC)$. Thus \[ \frac{PB}{BB_1} \cdot \frac{CC_1}{CQ} = \frac{T_AC}{T_AB} = \frac{CC_1}{BB_1} \]which implies $PB = CQ$ like we wanted.