Step 1: Write the distance rule.
The straight line distance from a point $(x,y)$ to a line $ax+by+c=0$ is $\dfrac{|ax+by+c|}{\sqrt{a^2+b^2}}$. Here both given lines have the same $\sqrt{a^2+b^2}=\sqrt{5}$, so the bottoms will cancel later.
Step 2: Turn the words into an equation.
The point $P$ is twice as far from the first line as from the second. So
\[ \frac{|2x-y+1|}{\sqrt5}=2\cdot\frac{|2x+y-2|}{\sqrt5}. \]
Step 3: Cancel the common bottom.
Both sides have $\sqrt5$ underneath, so we just drop it:
\[ |2x-y+1|=2\,|2x+y-2|. \]
Step 4: Test each option.
Instead of solving the whole locus, we just check which given point fits. This is fast and safe.
Step 5: Check the point $(1,1)$.
Left side: $|2(1)-1+1|=|2|=2$. Right side: $2|2(1)+1-2|=2|1|=2$. Both sides are equal.
Step 6: Confirm the others fail.
For example $(2,3)$ gives left $=2$ but right $=2|3|=6$, which are not equal. So only $(1,1)$ works.
\[ \boxed{(1,1)} \]