For line $M$, think about how its slope must relate to the slope of line $L$ if the lines are perpendicular. What is the negative reciprocal of $3$?

Once you know the slope of line $M$, plug it and the point $(-1,1)$ into point-slope form to get an equation for $M$. Then simplify to $y = mx + b$ form.

Set the two equations equal to each other to find the $x$-coordinate where they intersect. Then plug that $x$-value into either equation to find the $y$-coordinate of point $P$.

Type the simplified equation for line $L$, $y = 3x - 1$, into one of the expression lines in Desmos to graph it.

Use the perpendicular slope $-\tfrac{1}{3}$ and point $(-1,1)$ to get the equation $y = -\tfrac{1}{3}x + \tfrac{2}{3}$, then type this into another expression line in Desmos to graph line $M$.

Click on the point where the two lines cross; Desmos will show the coordinates of this intersection. Use the $y$-coordinate of that point as your answer.

Line $L$ passes through $(2,5)$ and has slope $3$.

Use point-slope form with point $(2,5)$:

Simplify to slope-intercept form:

So the equation of line $L$ is $y = 3x - 1$.

If two lines are perpendicular, their slopes are negative reciprocals: they multiply to $-1$.

Since line $L$ has slope $3$, line $M$ must have slope $-\dfrac{1}{3}$.

Line $M$ passes through $(-1,1)$, so use point-slope form:

Now simplify to slope-intercept form:

So the equation of line $M$ is $y = -\dfrac{1}{3}x + \dfrac{2}{3}$.

At the intersection point $P$, both lines have the same $x$ and $y$ values, so set the right sides of the equations equal:

Add $\dfrac{1}{3}x$ to both sides and add $1$ to both sides:

Solve for $x$ by multiplying both sides by $\dfrac{3}{10}$:

So the intersection point $P$ has $x$-coordinate $\dfrac{1}{2}$.

Substitute $x = \dfrac{1}{2}$ into either line's equation. Using line $L$, $y = 3x - 1$:

So the $y$-coordinate of $P$ is $\boxed{\dfrac{1}{2}}$, which corresponds to choice C.