Combinatorics using a geometric diagram How can I do this without trial-and-error? 

It has something to do with a triangle and summing the next row?
After seeing JorleArbake's comment linking to http://abelkonkurransen.no/problems/abel_1314_r1_sol_nb.pdf it includes a solution described here:



Their argument is to recursively find the number of paths from the starting cell by finding the number of paths starting from cells in the middle of the grid instead.

For all of the triangles on the bottom row, there is only one path starting from it and leading to the end (namely it itself).  From there, notice that picking a starting square on the next level up, you may continue downwards to any of the adjacent cells, thus inheriting all of their (possibly multiple) different paths having started from each of them.  Continuing this process working upwards, you find that you can find the total number of paths from some parent cell is equal to the sum of the total number of paths from each adjacent child.  Filling in the grid from bottom up leads eventually to $56+41+56 = 153$



-Edit- Below is my original answer to this question, which still has some merit in leaving up since it shows a method to at least bound the size on the answer using elementary arguments.

In a very common question similar to this one, the definition of "adjacent" would have been such that at every step we have only three moves possible (straight down, down-left, or down-right), in which case the answer would be $3^4=81$ seen quickly from the multiplication principle.

With the definition of "adjacency" given here as two cells are adjacent if their respective borders share at least one point in common (i.e., share a corner vertex or edge), you can quickly see that there will be at least 3 choices available at every step (and in particular at some steps will have 4 or even 5 choices available), and that the available paths from the previously mentioned problem will all be valid paths for this one.  As such we know that the answer will be strictly greater than 81.  This rules out the possibility of $(a)$ as being the correct answer.

Notice then also that there will be at most 5 choices available at each step, so the number of possibilities must be strictly less than $5^4=625$, thus ruling out the possibility of $(e)$ being the answer.

To proceed further, intuitively one might expect that instead of looking at $3^4$ and $5^4$, you could look at the average number of choices raised to the fourth power (through some argument relating to geometric mean).  The number of choices from each triangle in order is: 3,4,3,4,4,3,5,3,4,4,3,5,3,5,3,4.  So, the average seems slightly under 4.  Thus we would expect the total number of possibilities to be less than $4^4 = 256$, ruling out $(d)$ as being a correct answer.

This leaves us with two possibilities left, either $(b)$ or $(c)$.  I do not see at the moment a good argument to choose between these two.
Just want to point out that the average number of choices is $3.75$, so the answer must be lower than $197.75$. The answer must be b) $153$ (from choices provided).