Let $ABCD$ be a convex quadrilateral such that $AB=4$, $BC=5$, $CA=6$, and $\triangle{ABC}$ is similar to $\triangle{ACD}$. Let $P$ be a point on the extension of $DA$ past $A$ such that $\angle{BDC}=\angle{ACP}$. Compute $DP^2$.

[i]2020 CCA Math Bonanza Lightning Round #4.3[/i]
1. **Identify Similar Triangles:**
   Given that $\triangle ABC \sim \triangle ACD$, we can use the properties of similar triangles to find the lengths of $CD$ and $DA$.

2. **Calculate $CD$ and $DA$:**
   Since $\triangle ABC \sim \triangle ACD$, the ratios of corresponding sides are equal. Therefore, we have:
   \[
   \frac{AB}{AC} = \frac{AC}{AD} \quad \text{and} \quad \frac{BC}{CD} = \frac{AC}{AD}
   \]
   Given $AB = 4$, $BC = 5$, and $AC = 6$, we can set up the following proportion:
   \[
   \frac{4}{6} = \frac{6}{AD} \implies AD = \frac{6 \times 6}{4} = 9
   \]
   Similarly, for $CD$:
   \[
   \frac{5}{CD} = \frac{6}{9} \implies CD = \frac{5 \times 9}{6} = 7.5
   \]

3. **Use the Angle Condition:**
   Given $\angle BDC = \angle ACP$, we know that quadrilateral $BCDP$ is cyclic. This implies that $\angle BCP = \angle BDP$.

4. **Identify Similar Triangles Again:**
   Since $BCDP$ is cyclic, we have $\triangle BCD \sim \triangle PAC$. Using the similarity, we can find $PA$:
   \[
   \frac{BC}{CD} = \frac{PA}{AC} \implies \frac{5}{7.5} = \frac{PA}{6} \implies PA = \frac{5 \times 6}{7.5} = 4
   \]

5. **Calculate $DP$:**
   Since $P$ is on the extension of $DA$ past $A$, we have:
   \[
   DP = DA + PA = 9 + 4 = 13
   \]

6. **Compute $DP^2$:**
   \[
   DP^2 = 13^2 = 169
   \]

The final answer is $\boxed{169}$.