$ABCD$ - quadrilateral inscribed in circle, and $AB=BC,AD=3DC$ . Point $R$ is on the $BD$ and $DR=2RB$. Point $Q$ is on $AR$ and $\angle ADQ = \angle BDQ$.  Also $\angle ABQ + \angle CBD = \angle QBD$ . $AB$ intersect line $DQ$ in point $P$.
Find $\angle APD$
1. Given that $ABCD$ is a quadrilateral inscribed in a circle, we know that opposite angles of the quadrilateral sum up to $180^\circ$.
2. Given $AB = BC$ and $AD = 3DC$, we can infer that $\triangle ABC$ is isosceles with $AB = BC$.
3. Point $R$ is on $BD$ such that $DR = 2RB$. This implies $BR = \frac{1}{3}BD$ and $DR = \frac{2}{3}BD$.
4. Point $Q$ is on $AR$ such that $\angle ADQ = \angle BDQ$. This implies that $DQ$ is the angle bisector of $\angle BDA$.
5. Given $\angle ABQ + \angle CBD = \angle QBD$, we need to use this information to find $\angle APD$.

Let's proceed step-by-step:

1. **Identify the properties of the angle bisector:**
   Since $DQ$ is the angle bisector of $\angle BDA$, by the Angle Bisector Theorem, we have:
   \[
   \frac{BD}{DA} = \frac{BR}{RA}
   \]
   Given $AD = 3DC$, we can write $DA = 3DC$ and $BD = BR + DR = \frac{1}{3}BD + \frac{2}{3}BD = BD$.

2. **Use the given ratio $DR = 2RB$:**
   \[
   \frac{DR}{RB} = 2 \implies \frac{\frac{2}{3}BD}{\frac{1}{3}BD} = 2
   \]
   This confirms the given ratio.

3. **Use the given angle condition:**
   Given $\angle ABQ + \angle CBD = \angle QBD$, we need to use this to find $\angle APD$.

4. **Consider the isosceles triangle properties:**
   Since $AB = BC$, $\triangle ABC$ is isosceles. Let $M$ be the midpoint of $AD$ such that $AM = MN = ND$.

5. **Use the cyclic quadrilateral properties:**
   Since $ABCD$ is cyclic, $\angle BDA = \angle BCA$ and $\angle BAC = \angle CDE$.

6. **Analyze the triangle properties:**
   Since $\triangle CDE = \triangle NDE$, we have $CE = EN$.

7. **Use the angle bisector properties:**
   Since $\triangle BCD = \triangle BND$, we have $\angle CBD = \angle NBD$ and $BC = BN$.

8. **Analyze the bisector properties:**
   Since $\angle NBM = \angle MBA$, $BM$ is the bisector of $\angle NBA$.

9. **Use the cyclic properties:**
   Since $\angle BEC = 180^\circ - \angle CBE - \angle BCE = 180^\circ - \angle CAD - \angle BAC = 180^\circ - \angle BAD = 180^\circ - \angle BNA = \angle BND$, we have $\triangle BEC = \triangle ANE$.

10. **Conclude the properties:**
    Since $BE = NA$, $BR = MD$, and $\angle RBA = \angle MBA$, we have $BD = AD$.

11. **Use the altitude properties:**
    Since $DQ$ is the altitude of the isosceles triangle $ADB$, we have $\angle APD = 90^\circ$.

The final answer is $\boxed{90^\circ}$