There are differences in the construction and the angle chase for the cases if $P$ lies after point $I$ on $AI$, that is, the points are $A,I,P$ in that order on the line or in the order $A,P,I$ or in the order $P,A,I$. The ideas are very similar, but we'll at the different cases anyways (I've included diagrams in order to help see the cases more clearly). Firstly we'll consider the case $P,A,I$, that is, when $A\in PI$. Note that $AK=AN$, $BK=BL$, $CL=CM$, $DM=DN$ because in all these equalities the two segments are tangents to the circle inscribed in $ABCD$. Also, since $K,L,M,N$ are the touchpoints of the circle, let's call it $\Gamma$, with center $I$ and the sides $AB,BC,CD,DA$, this means that $IK\perp AB$, $IL\perp BC$, $IM\perp CD$, $IN\perp DA$. Combined with the equality of sides this gives that $\triangle AIK\cong \triangle AIN$, $\triangle BIK\cong \triangle BIL$, $\triangle CIL\cong\triangle CIM$, $\triangle DIM\cong \triangle DIN$. This also means that $N$ and $K$ are symmetric by the line $AI$, $K$ and $L$ are symmetric by the line $BI$, $L$ and $M$ are symmetric by the line $CI$, $M$ and $N$ are symmetric by the line $DI$. \newline\newline Denote $\angle IAK = \angle IAN=\alpha$, $\angle IBA=\angle IBL=\beta$, $\angle ICL=\angle ICM=\gamma$, $\angle MDI=\angle NDI=\delta$ and $\angle IPK=x$. From $\triangle AIB$, $\triangle BIC$, $\triangle CID$, $\triangle DIA$ we have that: $\angle AIB=180^{\circ}-\alpha-\beta$, $\angle BIC=180^{\circ}-\beta-\gamma$, $\angle CID=180^{\circ}-\gamma-\delta$ and $\angle DIA=180^{\circ}-\delta-\alpha$. Now in $\triangle PIQ$ we have that: $$\angle IQP=180^{\circ}-\angle IPQ-\angle PIQ=180^{\circ}-x-(180^{\circ}-\alpha-\beta)=\alpha+\beta-x$$We mentioned that $K$ and $L$ are symmetric wrt line $BI$, but $K\in BI\Longrightarrow \angle IQL=\angle IQK=\angle IQP=\alpha+\beta-x$. Now in $\triangle QIR$ we have that: $$\angle IRQ=180^{\circ}-\angle QIR-\angle IQR=180^{\circ}-(180^{\circ}-\beta-\gamma)-(\alpha+\beta-x)=\gamma-\alpha+x$$We mentioned that $L$ and $M$ are symmetric wrt line $CI$, but $R\in CI\Longrightarrow\angle MRI=\angle LRI=\angle IRQ=\gamma-\alpha+x$. Now in $\triangle RSI$ we have that: $$\angle ISR=180^{\circ}=\angle IRS-\angle SIR=180^{\circ}-(\gamma-\alpha+x)-(180^{\circ}-\gamma-\delta)=\delta+\alpha-x$$We mentioned that $M$ and $N$ are symmetric wrt line $DI$, but $S\in DI\Longrightarrow \angle NSI=\angle MSI=\angle ISR=\delta+\alpha-x$. We also mentioned that $N$ and $K$ are symmetric wrt line $AI$, but $P\in AI\Longrightarrow \angle NPI=\angle KPI=x$. We mentioned that $IN\perp DA\Longrightarrow \angle AIN=180^{\circ}-\angle IAN-\angle INA=180^{\circ}-\alpha-90^{\circ}=90^{\circ}-\alpha$. Also, $\angle SIN=\angle DIA-\angle AIN=(180^{\circ}-\delta-\alpha)-(90^{\circ}-\alpha)=90^{\circ}-\delta$. Therefore in $\triangle PNI$ we have that: $$\angle PNI=180^{\circ}-\angle NPI-\angle NIP=180^{\circ}-x-(90^{\circ}-\alpha)=90^{\circ}+\alpha-x$$In $\triangle SNI$ we have that: $$\angle SNI=180^{\circ}-\angle NSI-\angle SIN=180^{\circ}-(\delta+\alpha-x)-(90^{\circ}-\delta)=90^{\circ}-\alpha+x$$$$\Longrightarrow \angle SNI+\angle PNI=(90^{\circ}-\alpha+x)+(90^{\circ}+\alpha-x)=180^{\circ}\Longrightarrow P,N,S\text{ are collinear}$$Thus we've solved the problem because $\angle SNI+\angle PNI=180^{\circ}$ when we're in the case $P,A,I$. In the next cases we'll use the already made remarks, for example that $\angle KQI=\angle LQI$, $\angle LRI=\angle MRI$ etc. as these angle equalities are independent on the position of $P$. Also, we'll use the notation of angles as in this case, that is $\angle IAK = \angle IAN=\alpha$, $\angle IBA=\angle IBL=\beta$, $\angle ICL=\angle ICM=\gamma$, $\angle MDI=\angle NDI=\delta$ and $\angle IPK=x$. $\textbf{Case 2}$ $P\in AI$ $$\angle IQP=180^{\circ}-\angle IPK-\angle PIK=180^{\circ}-x-(180^{\circ}-\alpha-\beta)=\alpha+\beta-x$$$$\angle IRQ=180^{\circ}-\angle QIR-\angle IQR=180^{\circ}-(180^{\circ}-\beta-\gamma)-(\alpha+\beta-x)=\gamma-\alpha+x$$$$\angle ISR=180^{\circ}=\angle IRS-\angle SIR=180^{\circ}-(\gamma-\alpha+x)-(180^{\circ}-\gamma-\delta)=\delta+\alpha-x$$We have that $\angle NPI=\angle KPI=x$. We mentioned that $IN\perp DA\Longrightarrow \angle AIN=180^{\circ}-\angle IAN-\angle INA=180^{\circ}-\alpha-90^{\circ}=90^{\circ}-\alpha$. Also, $\angle SIN=\angle DIA-\angle AIN=(180^{\circ}-\delta-\alpha)-(90^{\circ}-\alpha)=90^{\circ}-\delta$. Therefore in $\triangle PNI$ we have that: $$\angle PNI=180^{\circ}-\angle NPI-\angle NIP=180^{\circ}-x-(90^{\circ}-\alpha)=90^{\circ}+\alpha-x$$In $\triangle SNI$ we have that: $$\angle SNI=180^{\circ}-\angle NSI-\angle SIN=180^{\circ}-(\delta+\alpha-x)-(90^{\circ}-\delta)=90^{\circ}-\alpha+x$$$$\Longrightarrow \angle SNI+\angle PNI=(90^{\circ}-\alpha+x)+(90^{\circ}+\alpha-x)=180^{\circ}\Longrightarrow P,N,S\text{ are collinear}$$$\textbf{Note:}$ This case is solved with the same angle chase as in the first case. $\textbf{Case 3}$ $I\in AP$, $I\in CR$. We have that: $$\angle QLB=\angle QKB=\angle PKB=\angle KAP+\angle KPA=\alpha+x$$$$\angle CRM=\angle CRL=\angle QLB-\angle RCL=(\alpha+x)-\gamma=\alpha-\gamma+x$$$$\angle DNS=\angle DMS=\angle MCR+\angle MRC=\gamma+(\alpha-\gamma+x)=\alpha+\gamma$$$$\angle DNP=\angle NAP+\angle NPA=\alpha+x$$$$\Longrightarrow \angle DNS=\angle DNP\Longrightarrow P,N,S\text{ are collinear}$$ $\textbf{Case 4}$ $I\in AP$, $C\in IR$. We have that: $$\angle QLB=\angle QKB=\angle KPA+\angle KAP=\alpha+x$$$$\angle MRI=\angle LRI=\angle ICL-\angle CLR=\gamma-\angle QLB=\gamma-\alpha-x$$$$\angle DNS=\angle DMS=\angle CMR=\angle MCI-\angle MRC=\gamma-(\gamma-\alpha-x)=\alpha+x$$$$\angle DNP=\angle NPA+\angle NAP=\angle APK+\alpha=x+\alpha$$$$\Longrightarrow \angle DNS=\alpha+x=\angle DNP\Longrightarrow P,N,S\textbf{ are collinear}$$ The case $I\in AP$, $R\in CI$ doesn't exist as if $R\in CI$, then $B\in IQ$, bu then $P\in AI$, contradiction, thus no such construction exists. $\textbf{Case 5}$ $P\equiv A$. In this case $Q\equiv B$, $C\equiv R$, $D\equiv S$, so $P,N,S$ are $A,N,D$ which are collinear as we know that $N$ lies on $AD$, because $N$ is the touchpoint of the inscribed circle with center $I$ to side $AD$. $\textbf{Case 6}$ $P\equiv I$. Here $I\equiv P\equiv Q\equiv R\equiv S$, so $P\equiv S$, so $P,N,S$ trivially lie on the line $IN$. Thus we've looked at all cases and proved all of them, so the problem is solved.

$\textbf{Complex bash solution}$ I've provided a solution above, but I just wanted to show a second one, which is achieved using complex numbers. It's more convenient, because we can avoid the different cases. Here is the solution:\newline\newline We'll use complex numbers. We'll denote the complex number of a point with the lower case letter. Let $i=0$, that is, $I$ is the origin of the complex plane and $(KLMN)$ be the unit circle. Then we have: $$\overline{k}=\frac{1}{k}\quad,\quad \overline{l}=\frac{1}{l}\quad,\quad \overline{m}=\frac{1}{m}\quad,\quad \overline{n}=\frac{1}{n}$$By the intersections of tangents to the circumcircle formula (that is, if $a,b$ lie on the unit circle, then the intersection of the tangents at $A$ and $B$ to the unit circle is given by $\frac{2ab}{a+b}$): $$\boxed{a=\frac{2nk}{n+k}}\quad, \quad \boxed{b=\frac{2kl}{k+l}}\quad, \quad \boxed{c=\frac{2ml}{m+l}}\quad, \quad \boxed{d=\frac{2mn}{m+n}}$$$$P\in AI\Longrightarrow \overline{p}=\frac{p}{nk}$$$$Q\in BI\Longrightarrow \overline{q}=\frac{q}{kl}$$$$R\in AI\Longrightarrow \overline{r}=\frac{r}{lm}$$$$S\in DI\Longrightarrow \overline{s}=\frac{s}{mn}$$$$Q\in RK\Longrightarrow (\overline{p}-\overline{k})q-(p-k)\overline{q}+p\overline{k}-\overline{p}k=0$$$$\Longrightarrow \left(\frac{p}{nk}-\frac{1}{k}\right)q-(p-k)\frac{q}{kl}+\frac{p}{k}-\frac{p}{n}=0$$$$\Longrightarrow (pl-nl)q-(p-k)nq+pnl-pkl=0$$$$\Longrightarrow \boxed{q=\frac{pkl-pnl}{-nl-pn+pl+kn}}$$$$R\in QL\Longrightarrow (\overline{q}-\overline{l})r-(q-l)\overline{r}+q\overline{l}-\overline{q}l=0$$$$\Longrightarrow \left(\frac{q}{kl}-\frac{1}{l}\right)r-(q-l)\frac{r}{lm}+\frac{q}{l}-\frac{q}{k}=0$$$$\Longrightarrow (qm-km)r-(q-l)kr+qkm-qlm=0$$$$\Longrightarrow \boxed{r=\frac{qlm-qkm}{-km-qk+qm+kl}}$$$$S\in RM\Longrightarrow \boxed{s=\frac{rn(m-l)}{-ln-rl+rn+lm}}$$$$S\in PN\iff (\overline{p}-\overline{n})s-(p-n)\overline{s}+p\overline{n}-\overline{p}n=0$$$$\iff \left(\frac{p}{nk}-\frac{1}{n}\right)s-(p-n)\frac{s}{mn}+\frac{p}{n}-\frac{p}{k}=0$$$$\iff (mp-mk)s-(pk-nk)s+pmk-pmn=0$$$$\iff s=\frac{pmn-pmk}{-mk-pk+mp+nk}$$$$\iff \frac{pmn-pmk}{-mk-pk+mp+nk}=\frac{rn(m-l)}{-ln-rl+rn+lm}$$$$\iff (pmn-pmk)(ln+rl-rn-lm)=(rmn-rnl)(mk+pk-mp-nk)$$$$\iff (m-n)(klmp+klnr-klpr-kmnr-lmnp+mnpr)=0$$$$\iff r=\frac{lmnp-klmp}{kln-klp-kmn+mnp}$$$$\iff \frac{qlm-qkm}{-km-qk+qm+kl}=\frac{lmnp-klmp}{kln-klp-kmn+mnp}$$$$\iff \frac{ql-qk}{-km-qk+qm+kl}=\frac{lnp-klp}{kln-klp-kmn+mnp}$$$$\iff k(l-m)(klp-knq-lnp+lnq-lpq+npq)=0$$$$\iff q=\frac{klp-lnp}{kn-ln+lp-np}$$However, we got exactly that for $q$, so $P,N,S$ are collinear and the problem is solved.