For the quadratic equation $ax^2+bx+c=0$ to have repeated roots, its discriminant must be zero: $b^2-4ac=0$ So we need $b^2=4ac.$

The coefficients $a,b,c$ are chosen from the set $\{1,2,3,4,5,6,7,8\}.$ Since each of $a,b,c$ can be chosen independently in $8$ ways, the total number of possible equations is $8^3=512.$

Now count the number of triples $(a,b,c)$ such that $b^2=4ac.$ This means $b$ must be even, because $b^2$ is divisible by $4$.

Possible even values of $b$ are: $b=2,4,6,8.$

Check each possible $b$:

(i) $b=2$ $b^2=4 \implies 4ac=4 \implies ac=1$ Since $a,c\in\{1,2,\dots,8\}$, this gives only $(a,c)=(1,1).$ So number of solutions = $1$.

(ii) $b=4$ $b^2=16 \implies 4ac=16 \implies ac=4$ Possible ordered pairs $(a,c)$ are: $(1,4),(2,2),(4,1).$ So number of solutions = $3$.

(iii) $b=6$ $b^2=36 \implies 4ac=36 \implies ac=9$ Possible ordered pairs in the set are: $(1,9),(3,3),(9,1).$ But only $(3,3)$ is allowed since values must lie in $\{1,2,\dots,8\}$. So number of solutions = $1$.

(iv) $b=8$ $b^2=64 \implies 4ac=64 \implies ac=16$ Possible ordered pairs $(a,c)$ in the set are: $(2,8),(4,4),(8,2).$ So number of solutions = $3$.

Total favorable cases: $1+3+1+3=8.$

Therefore, the required probability is $\frac{8}{512}=\frac{1}{64}.$

Comparing with the stored correct answer: Stored answer = B: $\frac{1}{64}$. This matches our result.