Let $ (X) \equiv \odot(AA_1P)$ cut $ DA$ again at $ A_2$ and let $ (Y) \equiv \odot (CC_1P)$ cut $ CD$ again at $ C_2.$ $ \angle CQP = \angle PC_1B = \angle PAA_2 = \pi - \angle A_2QP$ $ \Longrightarrow$ $ C, Q, A_2$ are collinear and similarly, $ A, Q, C_2$ are collinear. From $ (X)$, $ A_1Q$ is antiparallel of $ AA_2 \parallel BC$ WRT angle $ \angle (AB, A_2C)$ $ \Longrightarrow$ $ A_1BCQ$ is cyclic, and $ A_2P$ is antiparallel of $ AA_1 \parallel DC$ WRT angle $ \angle (DA, CA_1)$ $ \Longrightarrow$ $ CDA_2P$ is also cyclic. From $ \odot (A_1BCQ), \odot(CDA_2P),$ $ \angle QBA = \angle QBA_1 = \angle QCA_1 = \angle QCP = \angle A_2CP = \angle A_2DP = \angle ADP.$

From Virgil Nicula's extension we have $ \frac {\sin \angle ABQ}{\sin \angle CBQ} = \frac {AA_1}{CC_1}$ Denote by $ E$ the intersection of the lines $ AD$ and $ CP$. Then $ \dfrac{\sin \angle ADP}{\sin \angle CDP} = \dfrac{EP}{PC} \cdot \dfrac{CD}{ED}$ But $ \dfrac{EP}{PC} = \dfrac{EA}{CC_1}$ and $ \dfrac{AD}{ED} = 1 - \dfrac{EA}{ED} = 1 - \dfrac{AA_1}{CD} = \dfrac{CD - AA_1}{CD}$ $ ED = \dfrac{CD \cdot AD}{CD - AA_1} = \dfrac{CD \cdot AD}{AB - AA_1} = \dfrac{CD \cdot AD}{BA_1}$ So $ \dfrac{\sin \angle ADP}{\sin \angle CDP} = \dfrac{EA}{CC_1} \cdot \dfrac{CD}{\dfrac{CD \cdot AD}{BA_1}} = \dfrac{EA}{BC} \cdot \dfrac{BA_1}{CC_1} = \dfrac{AA_1}{BA_1} \cdot \dfrac{BA_1}{CC_1} = \frac {AA_1}{CC_1} = \frac {\sin \angle ABQ}{\sin \angle CBQ}$ As function $ f(x) = \dfrac{\sin (\alpha - x)}{\sin x}$ is strictly decreasing on $ (0,\pi)$ we get that $ \angle PDA = \angle QBA$.

Sorry to revive this old thread, But I have an alternative solution to this beautifull problem.

Use complex numbers. Let $Q=0$. Note that in angle measure we have $QAP=QA_1P=QA_1C=QBC$, $QPA=QA_1A=QCC_1=QCB$ since $Q$ takes $AC_1$ to $A_1C$. So $Q$ takes $AP$ to $BC$ and $pb=ac$. So $p(a+c-d)=ac$, $-pd=-ap-cp+ac$, and $p(p-d)=(p-a)(p-c)$. Thus, $P$ takes $AD$ to $QC$. So now $PDA=PCA=A_1CQ=A_1BQ=QBA$, so we're done.

Nice angle chasing exercise! [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.182cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 15.96, ymin = -5.24, ymax = 6.3; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen bfffqq = rgb(0.7490196078431373,1.,0.); pen ffqqff = rgb(1.,0.,1.); pen xfqqff = rgb(0.4980392156862745,0.,1.); /* draw figures */ draw((4.32,3.2)--(5.36,0.), xdxdff); draw((2.66,0.)--(5.36,0.), xdxdff); draw((1.62,3.2)--(4.32,3.2), xdxdff); draw((1.62,3.2)--(2.66,0.), xdxdff); draw((4.32,3.2)--(4.1,0.)); draw((4.859378365410247,1.5403742602761632)--(2.66,0.)); draw(circle((3.38,0.5817777444167372), 0.9256702133582054), linetype("2 2") + ffxfqq); draw(circle((3.919179933099226,2.152271624016165), 1.1217804936225562), linetype("2 2") + ffxfqq); draw(circle((4.73,1.56425), 1.686350515907057), linetype("4 4") + bfffqq); draw((3.518359866198457,3.2)--(2.66,0.), linewidth(1.2) + linetype("4 4") + ffqqff); draw(circle((2.569179933099227,1.739483478257249), 1.7418527652134657), dotted + xfqqff); /* dots and labels */ dot((2.66,0.),dotstyle); label("$A$", (2.38,-0.26), NE * labelscalefactor); dot((5.36,0.),dotstyle); label("$B$", (5.46,-0.38), NE * labelscalefactor); dot((4.32,3.2),dotstyle); label("$C$", (4.38,3.36), NE * labelscalefactor); dot((1.62,3.2),linewidth(3.pt) + dotstyle); label("$D$", (1.36,3.04), NE * labelscalefactor); dot((4.1,0.),dotstyle); label("$A_1$", (4.14,-0.44), NE * labelscalefactor); dot((4.859378365410247,1.5403742602761632),dotstyle); label("$C_1$", (5.02,1.52), NE * labelscalefactor); dot((4.172843888664785,1.059547471487769),linewidth(3.pt) + dotstyle); label("$P$", (4.34,0.78), NE * labelscalefactor); dot((4.172843888664785,1.059547471487769),linewidth(3.pt) + dotstyle); dot((3.0478152622528927,1.4457908484299304),linewidth(3.pt) + dotstyle); label("$Q$", (3.12,1.56), NE * labelscalefactor); dot((4.32,3.2),linewidth(3.pt) + dotstyle); dot((3.518359866198457,3.2),linewidth(3.pt) + dotstyle); label("$A_2$", (3.6,3.32), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Lemma 1. Points $A_1, B, C, Q$ lie on a circle. Proof. We have $$\angle A_1QC=\angle A_1QP+\angle PQC=\angle A_1AP+\angle PC_1B=180^{\circ}-\angle A_1BC_1 \Longrightarrow A_1, B, C, Q \, \text{are concyclic}. \, \square$$ Lemma 2. Points $A, Q, A_2$ are collinear where $A_2$ is the second intersection of $(CC_1P)$ with $CD$. Proof. Note that from $A_2C \parallel A_1B$ we have $$\angle AQP+\angle A_2QP=\angle PA_1B+\left(180^{\circ}-\angle PCA_2\right)=180^{\circ} \Longrightarrow A, Q, A_2 \, \text{are collinear}. \, \square$$ Lemma 3. Points $D, A_2, P, A$ are concyclic. Proof. We have $$\angle APA_2=\angle A_2CC_1=180^{\circ}-\angle ADA_1 \Longrightarrow A, D, A_2, P \, \text{are concyclic}. \, \square$$ As $$\angle PDA=\angle PA_2Q=\angle QCA_1=\angle QBA,$$we are done.

Let $APQ$ interrsect $AD$ at $X$ and let $CPQ$ intersect $CD$ at $Y$. Observe that $\angle XAP = \angle PC_1B = \angle PQC$. Thus $X, Q, C$ are collinear. Similarly, $A, Q, Y$ are collinear. This mean that $DAPY$ is cyclic and $BA_1QC$ is cyclic (properties of the Miquel Point). This gives us $\angle PDA = \angle PYA = \angle PYQ = \angle PCQ = \angle A_1CQ = \angle ABQ$ as required.

Pick $A_2 \in \overline{AD}, C_2 \in \overline{CD}$ such that $AA_1A_2PQ$, $CC_1C_2PQ$ are concylic. Observe that $Q$ is the Miquel point of $PA_1BC_1$. $\angle PQA_2 + \angle PQC = 180 - \angle PAA_2 + 180 - \angle PC_1C = 180$, and hence $A_2, Q, C$ are colinear. Similarly $C_2, A, Q$ are colinear. Thus $P$ is the Miquel point of $QC_2DA_2$. Now we're done since $\angle QBA = \angle QC_1A = \angle PC_1Q = \angle PC_2Q = \angle AC_2Q = \angle ADP$

Let circle $AA_1P$ and $CC_1P$ meet $AD$ and $CD$ at $S,K$. Claim $: BA_1QC$ is cyclic. Proof $:$ Note that $\angle CQA_1 = \angle CQP + \angle PQA_1 = \angle BC_1A + \angle PAB = \angle 180 - \angle A_1BC$. Claim $: A,Q,K$ and $C,Q,S$ are collinear. Proof $:$ Note that $\angle PQA = \angle PA_1B = \angle PCK = \angle 180 - \angle PQK$. we prove the other one with same approach. Claim $: CPSD$ is cyclic. Proof $:$ Note that $\angle PSA = \angle PQA = \angle PCK = \angle PCD$. Now Note that $\angle QBA = \angle QCA_1 = \angle QCP = \angle SCP = \angle SDP = \angle PDA$. we're Done.

Denote by $R$ reflection of $P$ wrt center of parallelogram. In $A_1BC_1P$ Miquel point $Q$ is the isogonal conjugate of $\infty_{BR},$ since $BR$ is homothetic to it's Gauss line with center $P$ and coefficient $2,$ therefore $$\angle PDA\stackrel{\text{symmetry}}{=}\angle RBC=\angle QBA.$$