Given $4a_n=n^2+5n+6=(n+2)(n+3)$ Hence, $a_n=\frac{(n+2)(n+3)}{4}$

Find $\dfrac{1}{a_k}$: $\frac{1}{a_k}=\frac{4}{(k+2)(k+3)}$

Therefore, $S_n=\sum_{k=1}^n \frac{4}{(k+2)(k+3)}$

Use partial fractions: $\frac{4}{(k+2)(k+3)}=4\left(\frac{1}{k+2}-\frac{1}{k+3}\right)$ since $\frac{1}{k+2}-\frac{1}{k+3}=\frac{1}{(k+2)(k+3)}$

So the sum telescopes: $S_n=4\sum_{k=1}^n \left(\frac{1}{k+2}-\frac{1}{k+3}\right)$

Expanding a few terms, $S_n=4\left(\frac13-\frac14+\frac14-\frac15+\frac15-\frac16+\cdots+\frac{1}{n+2}-\frac{1}{n+3}\right)$

All middle terms cancel, giving $S_n=4\left(\frac13-\frac{1}{n+3}\right)$

For $n=2025$, $S_{2025}=4\left(\frac13-\frac{1}{2028}\right)$

Now, $\frac13-\frac{1}{2028}=\frac{2028-3}{6084}=\frac{2025}{6084}=\frac{675}{2028}$

Hence, $S_{2025}=4\cdot \frac{675}{2028}=\frac{2700}{2028}=\frac{225}{169}$

Now compute: $507S_{2025}=507\cdot \frac{225}{169}$

Since $507=3\cdot 169$, $507S_{2025}=3\cdot 225=675$

Therefore, the correct option is: $\boxed{675}$ which is Option B.