The probability is
$\frac13$
Explanation :
The first moves are symmetrical, so let's consider that the kangaroo moves to B. Let the probability that the walk ends in D be $p$.
As the kangaroo is in B, the walk ends in D if he goes to C. If he goes back to A, then he has to jump back to B, at which point the probability of ending in D is $p$ again. Therefore $p = \frac12 + \frac{p}{4}$, or $p = \frac23$. As we're looking for the probability of ending in C, it should be equal to $1-p$.
This is not a coincidence. Generalizing for a polygon with any number of vertices:
For every vertex, there is a point in which the kangaroo first reaches an adjacent corner. Then that vertex is the last if and only if the kangaroo proceeds to reach the vertex by going around the long way. The probability of that happening is the same for every vertex, so the probability that the walk ends in every vertex is the same (except for the sad starting vertex).