Let $d$ touch $(O),(F)$ at $U,V.$ $R$ is the second intersection of $(O),(F)$ and $M \equiv AR \cap OF$ is the midpoint of $\overline{PQ}.$ $K$ and $J \equiv UV \cap OF$ are the insimilicenter and exsimilicenter of $(O) \sim (F),$ respectively. Since $(B,C,D,E)=-1,$ then $(O) \perp (F),$ thus if $UF$ cuts $(O)$ and $(F)$ again at $T$ and $X,Y,$ we have $(U,T,X,Y)=-1$ $\Longrightarrow$ $VT$ is the polar of $U$ WRT $(F)$ $\Longrightarrow$ $VT \perp FU$ $\Longrightarrow$ $VT$ goes through the antipode $U^*$ of $U$ WRT $(O)$ $\Longrightarrow$ $K \in VT.$ Since $U(J,K,O,F)=-1,$ then $(V,K,U^*,T)=-1$ $\Longrightarrow$ $VP \perp OF$ is the polar of $K$ WRT $(O)$ $\Longrightarrow$ $OA^2=OK \cdot OP$ $\Longrightarrow$ $\angle APK=\angle OAK.$ But $\frac{_{OA}}{^{FA}}=\frac{_{KO}}{^{KF}}$ implies that $AK$ bisects $\angle OAF,$ i.e. $\angle OAK=45^{\circ},$ so $\angle APK=45^{\circ}$ $\Longrightarrow$ the right $\triangle MPA$ is isosceles with $MA=MP,$ i.e. $PQ=m.$

WLOG $AB > AC$. $\angle FAC = \angle FAD - \angle CAD = \angle ADF - \angle DAB = \angle ABC$ $\Longrightarrow$ $AF$ is tangent of $(O)$ $\Longrightarrow$ $AO \perp AF$. The common tangent touches $(O), (F)$ at $U, V$. $PQ = \frac{UV^2}{OF} = \frac{OF^2 - (AO - AF)^2}{OF} = \frac{2 \cdot AO \cdot AF}{OF} = m$.