Is it specifically a DAG? If so, process nodes in topological order, keeping count of how many different paths end at each node.
The stationary measure will be the uniform measure, which you could see by a soft argument: the transition matrix is irreducible and aperiodic, and thus has a stationary measure; furthermore, it is invariant under the map $k \mapsto k+1$, implying that each state must get the same measure.
You could also show this manually: let $\mu_j = \frac{1}{N+1}$ for each $j \in \{0,1,\ldots,N\}$. Then $$(P \mu)_j = p \mu_{j-1} + r \mu_{j} + q \mu_{j+1} = \frac{p + r + q}{N+1} = \frac{1}{N+1} = \mu_j$$
where the indices are $\mod{N+1}$.
EDIT: some further comment on the invariance thing; let $A$ be the map that cycles the states, i.e. $A (x_0 ,\ldots, x_{N})^T = (x_N,x_0,x_1,\ldots,x_{N-1})^T\,.$
Then note that $PA = P$. If you set $\mu$ to be the stationary measure of $P$ (i.e. $\mu P = \mu$), then it must be invariant under $A$: $$\mu PA = \mu P \implies \mu A = \mu\,.$$ Iteratively applying $A$ shows that $\mu$ must be uniform.
Best Answer
If I understand the problem correctly, you need to construct a transition probability matrix $P=(p_{ij})$ where the probability $p_{ij}$ of transitioning from vertex $i$ to vertex $j$ is $1/\mathrm{deg}^{\mathrm{out}}(i)$, where $\mathrm{deg}^{\mathrm{out}}(i)$ denotes the out degree of vertex $i$. One you have $P$, the stationary distribution (if it exists) is the eigenvector corresponding to eigenvalue $1$; the eigenvector is normalised so that it sums to $1$.
Here's a simple worked example. For the graph:
the transition probability matrix is $$P=\left(\begin{array}{ccc} 0 & 1 & 0 \\ \tfrac{1}{2} & 0 & \tfrac{1}{2} \\ 1 & 0 & 0 \\ \end{array}\right).$$ Here the stationary distribution is the eigenvector $$\vec{v}=(\tfrac{2}{5},\tfrac{2}{5},\tfrac{1}{5})$$ since $\vec{v}P=\vec{v}$.