Let $ABCD$ be a convex quadrilateral with $\angle DAB =\angle B DC = 90^o$. Let the incircles of triangles $ABD$ and $BCD$ touch $BD$ at $P$ and $Q$, respectively, with $P$ lying in between $B$ and $Q$. If $AD = 999$ and $PQ = 200$ then what is the sum of the radii of the incircles of triangles $ABD$ and $BDC$ ?
1. **Label the tangent points and define variables:**
   - Let the tangent of side \( AD \) to the incircle of \(\triangle ABD\) be \( F \).
   - Let \( QD = x \).
   - Since \( PQ = 200 \), we have \( PD = 200 + x \).

2. **Use the property of tangents from a point to a circle:**
   - By the property of tangents from a point to a circle, \( PD = FD \).
   - Therefore, \( FD = 200 + x \).

3. **Express \( AF \) in terms of \( x \):**
   - Since \( AD = 999 \) and \( AF + FD = AD \), we have:
     \[
     AF + (200 + x) = 999
     \]
   - Solving for \( AF \):
     \[
     AF = 999 - 200 - x = 799 - x
     \]

4. **Identify the inradii of the triangles:**
   - Note that \( AF \) is the inradius of \(\triangle ABD\) since it is a right triangle.
   - Similarly, \( QD = x \) is the inradius of \(\triangle BDC\) because it is also a right triangle.

5. **Sum the inradii:**
   - The sum of the radii of the incircles of \(\triangle ABD\) and \(\triangle BDC\) is:
     \[
     AF + QD = (799 - x) + x = 799
     \]

The final answer is \(\boxed{799}\).