Let the parallel to $AD$ through $P$ intersect $BD$ at $O$. We show that $O$ is the circumcenter of $DPQ$.

One has $\vec{ADP} = \vec{BDP}$ ($DP$ bisects $\vec{ADB}$) and $\vec{ADP} = \vec{OPD}$ (as $PO \parallel AD$). Hence $\vec{ODP} = \vec{OPD}$ and so $OP = OD$. Also $DO = AP$, $BO = BP$ as triangle $ADB$ is isosceles, then $\frac{DO}{BO} = \frac{AP}{BP} = \frac{DA}{DB} = \frac{b}{a}$ by the bisector theorem in triangle $ABD$.

On the other hand the same theorem in triangle $BCD$ gives $\frac{CQ}{BO} = \frac{DC}{DB} = \frac{b}{a}$. In summary,

$$\frac{DO}{BO} = \frac{b}{a} = \frac{CQ}{BO}.$$

Now the converse of Thales' theorem implies $QQ \parallel CD$. Then

$B\hat{DQ} = C\hat{DQ} = O\hat{QD}$, hence $OQ = OD$. We obtained $OP = OQ = OD$, meaning that $O$ is the circumcenter of $DPQ$.

In addition $OD = AP$ from the isosceles triangle $ADB$. Since $AP + BP = a$ and $\frac{AP}{BP} = \frac{b}{a}$, standard calculus lead to $AP = \frac{ab}{a+b}$, so triangle $DPQ$ has circumradius $\frac{ab}{a+b}$.