Step 1: Tidy the hyperbola.
Take $x^2-4y^2-2x-8y-19=0$ and complete the squares:
\[ (x-1)^2-4(y+1)^2-16=0. \]
Step 2: Write the standard form.
\[ \frac{(x-1)^2}{16}-\frac{(y+1)^2}{4}=1. \]
Step 3: Recall the conjugate rule.
The conjugate hyperbola uses the same centre and terms but the right side becomes $-1$ instead of $+1$.
Step 4: Apply the rule.
\[ \frac{(x-1)^2}{16}-\frac{(y+1)^2}{4}=-1\ \Rightarrow\ (x-1)^2-4(y+1)^2=-16. \]
Step 5: Expand back out.
$(x^2-2x+1)-4(y^2+2y+1)+16=0$, which becomes
\[ x^2-4y^2-2x-8y+13=0. \]
Step 6: Pick the matching option.
The constant becomes $+13$, so the conjugate hyperbola is $x^2-4y^2-2x-8y+13=0$.
\[ \boxed{x^2-4y^2-2x-8y+13=0} \]