Set coordinates of points as $A,B,C,D=(-a,0),(b,0),(0,c),(-d,0)$ where $a,b,c,d>0$ and $P=(0,0)$. Then $Q=(c-a,0)$ Perimeter of $BDQ$ $= \sqrt{b^2+(c-a)^2}+\sqrt{(c-a)^2+d^2}+(b+d) \geq \sqrt{(b+d)^2+4(c-a)^2}+(b+d)$ We will prove, that $\sqrt{(b+d)^2+2(c-a)^2}+(b+d) \geq 2AC=2(a+c)$ or $\sqrt{(b+d)^2+2(c-a)^2} \geq 2(a+c)-(b+d)$ $4(c-a)^2 \geq 4(a+c)^2-4(a+c)(b+d)$ $(a+c)(b+d) \geq 4ac$ As $ABCD$ is cyclic, then $\frac{b}{a}=\frac{c}{d}= \tan BAC=x$ So $(a+c)(b+d)=(a+c)(ax+\frac{c}{x}) \geq 4ac$

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lemma Let $ B,C,U,V \ 4 $ points s.t. $U$ is on the $BC$ bisector and $UV \parallel BC$ then $BV+VC\ge BU+UC $ proof : let $C'$ the symmetric of $C$ in $UV$ then $B,U,C'$ are collinear and $BV+VC=BV+VC'\ge BC'=BU+UC'=BU+UC$ Back to the problem Let $O,I$ the circumcenter of $ABCD$, the midpoint of $AC$ ;$B',D' $the symmetrics of $ B,D $ in $I$ $BQ+BP\ge BB',DP+DQ\ge DD'$ so the perimeter $\ge 2( BI+ID)\overset{\text{by Lemma}}{\ge}2( BO+OD)$ since $IO \parallel BD$ therefore we get the desired result

Let's note $PQ=x$ , $AP=QC=a$ , $BP=b$ and $DP=c$. Because $ABCD$ is cyclic we have $BP*PD=AP*PC$ , so $bc=a(x+a)$.The perimeter of triangle $BDQ$ is $\sqrt{b^2+x^2} + \sqrt{c^2+x^2} + b+c $.But we have $a^2+ax-bc=0$ , so $a=\frac{-x\pm \sqrt{x^2+4bc}}{2}$ . Because $a$ is the length of $AP$ , $a$ is a positive number , so $a=\frac{-x+\sqrt{x^2+4bc}}{2}$. We must prove this inequality$\sqrt{b^2+x^2} + \sqrt{c^2+x^2} + b+c >= 2(2a+x) $ which is equivalent to $\sqrt{b^2+x^2} + \sqrt{c^2+x^2} + b+c >=\sqrt{x^2+4bc}$ . By Minkowski inequality we have $\sqrt{(b+c)^2 +4x^2} +b+c>=2\sqrt{x^2+4bc}$ Square it and after divide by $2$ we obtain $(b+c)^2+(b+c)\sqrt{(b+c)^2+4x^2}>=8bc$ which is true because $(b+c)^2>=4bc$ and $(b+c)\sqrt{(b+c)^2+4x^2}>=(b+c)(b+c)>=4bc$. Equality when $x=0$ , $P=Q$ so when $ABCD$ is a square.