Let $ \triangle ABE $ be a triangle with $ \frac{AB}{3} = \frac{BE}{4} = \frac{EA}{5} $. Let $ D \neq A $ be on line $ \overline{AE} $ such that $ AE = ED $ and $ D $ is closer to $ E $ than to $ A $. Moreover, let $ C $ be a point such that $ BCDE $ is a parallelogram. Furthermore, let $ M $ be on line $ \overline{CD} $ such that $ \overline{AM} $ bisects $ \angle BAE $, and let $ P $ be the intersection of $ \overline{AM} $ and $ \overline{BE} $. Compute the ratio of $ PM $ to the perimeter of $ \triangle ABE $.
1. **Define Variables and Ratios:**
   Let \( AB = 3k \), \( BE = 4k \), and \( EA = 5k \) for some constant \( k \). This is derived from the given ratio \( \frac{AB}{3} = \frac{BE}{4} = \frac{EA}{5} \).

2. **Position of Point \( D \):**
   Since \( D \) is on line \( \overline{AE} \) such that \( AE = ED \), we have \( ED = 5k \). Therefore, \( AD = AE + ED = 5k + 5k = 10k \).

3. **Parallelogram \( BCDE \):**
   Since \( BCDE \) is a parallelogram, \( BC = DE = 5k \) and \( CD = BE = 4k \).

4. **Angle Bisector \( AM \):**
   The line \( AM \) bisects \( \angle BAE \). By the Angle Bisector Theorem, the ratio \( \frac{BP}{PE} = \frac{AB}{AE} = \frac{3k}{5k} = \frac{3}{5} \).

5. **Intersection Point \( P \):**
   Let \( P \) be the intersection of \( \overline{AM} \) and \( \overline{BE} \). Using the Angle Bisector Theorem, we can find the lengths \( BP \) and \( PE \):
   \[
   BP = \frac{3k \cdot 4k}{3k + 5k} = \frac{12k^2}{8k} = \frac{12k^2}{8k} = \frac{3k}{2}
   \]
   \[
   PE = \frac{5k \cdot 4k}{3k + 5k} = \frac{20k^2}{8k} = \frac{20k^2}{8k} = \frac{5k}{2}
   \]

6. **Length of \( PM \):**
   Since \( PM \) is half the length of the \( A \)-angle bisector in \( \triangle AFD \) and \( \triangle ABE \sim \triangle AFD \) in a \( 1:2 \) ratio, we need to find the length of the angle bisector in \( \triangle ABE \).

7. **Using Stewart's Theorem:**
   Stewart's Theorem for \( \triangle ABE \) with angle bisector \( AM \) gives:
   \[
   AB^2 \cdot PE + BE^2 \cdot BP = AE \cdot (AP^2 + PM^2)
   \]
   Substituting the known values:
   \[
   (3k)^2 \cdot \frac{5k}{2} + (4k)^2 \cdot \frac{3k}{2} = 5k \cdot (AP^2 + PM^2)
   \]
   \[
   9k^2 \cdot \frac{5k}{2} + 16k^2 \cdot \frac{3k}{2} = 5k \cdot (AP^2 + PM^2)
   \]
   \[
   \frac{45k^3}{2} + \frac{48k^3}{2} = 5k \cdot (AP^2 + PM^2)
   \]
   \[
   \frac{93k^3}{2} = 5k \cdot (AP^2 + PM^2)
   \]
   \[
   \frac{93k^2}{10} = AP^2 + PM^2
   \]

8. **Perimeter of \( \triangle ABE \):**
   The perimeter of \( \triangle ABE \) is:
   \[
   AB + BE + EA = 3k + 4k + 5k = 12k
   \]

9. **Ratio of \( PM \) to Perimeter:**
   Since \( PM \) is half the length of the angle bisector in \( \triangle AFD \), and the perimeter of \( \triangle ABE \) is \( 12k \), the ratio \( \frac{PM}{\text{Perimeter}} \) is:
   \[
   \frac{PM}{12k}
   \]

The final answer is \(\boxed{\frac{PM}{12k}}\)