Let $ ABCD$ be a trapezoid inscribed in a circle $ \mathcal{C}$ with $ AB\parallel CD$, $ AB\equal{}2CD$. Let $ \{Q\}\equal{}AD\cap BC$ and let $ P$ be the intersection of tangents to $ \mathcal{C}$ at $ B$ and $ D$. Calculate the area of the quadrilateral $ ABPQ$ in terms of the area of the triangle $ PDQ$.
1. **Define the given values and relationships:**
   - Let \( AB = 2a \) and \( CD = a \).
   - Since \( AB \parallel CD \), \( ABCD \) is a trapezoid.
   - Let \( AD = b \) and \( BC = b \).
   - Let \( S_{\triangle PDQ} = x \) be the area of triangle \( PDQ \).
   - Let \( h \) be the altitude of triangle \( QAD \).

2. **Establish the similarity and area relationships:**
   - Since \( \angle CBP = \angle QDP \) and \( CB = DQ \), triangles \( \triangle QDP \) and \( \triangle PCB \) are congruent.
   - Therefore, \( S_{\triangle QDP} = S_{\triangle PCB} \).
   - Similarly, \( S_{\triangle QPA} = S_{\triangle QPB} \), implying that \( ABPQ \) is a trapezoid.

3. **Determine the length of \( PQ \):**
   - Let \( R \) be the intersection of \( DP \) with \( AB \).
   - By similarity of triangles \( \triangle RAD \) and \( \triangle ADC \), we have:
     \[
     \triangle RAD \sim \triangle ADC \implies \frac{RA}{AD} = \frac{AD}{DC} \implies \frac{RA}{b} = \frac{b}{a} \implies RA = \frac{b^2}{a}
     \]
   - Since \( PQ = RA \), we have:
     \[
     PQ = \frac{b^2}{a}
     \]

4. **Calculate the altitude \( h \) of triangle \( QAD \):**
   - Using the area formula for triangle \( PQA \):
     \[
     S_{\triangle PQA} = 2x = \frac{h \cdot PQ}{2} \implies h = \frac{4x}{PQ} = \frac{4ax}{b^2}
     \]

5. **Calculate the area of quadrilateral \( ABPQ \):**
   - The area of \( ABPQ \) can be split into two parts: \( S_{ABQ} \) and \( S_{BPQ} \).
   - The area of \( S_{ABQ} \) is:
     \[
     S_{ABQ} = \frac{h \cdot 2a}{2} = \frac{4ax \cdot 2a}{2 \cdot b^2} = \frac{4a^2x}{b^2}
     \]
   - The area of \( S_{BPQ} \) is:
     \[
     S_{BPQ} = 2x
     \]
   - Therefore, the total area of \( ABPQ \) is:
     \[
     S_{ABPQ} = S_{ABQ} + S_{BPQ} = \frac{4a^2x}{b^2} + 2x
     \]

The final answer is \( \boxed{ \frac{4a^2x}{b^2} + 2x } \)