Conjecture related to abc-conjecture, with $a1$ Let $\operatorname{rad}(b)$ be the product of all distinct prime factors of $b$. The numbers $\,a,b,c\,$ is a $abc$-triple if they are coprime and $a+b=c$. One version of the abc-conjecture is then:

For all $\varepsilon>0$ the set $E_\varepsilon$ of all $abc$-triplets with 
  $c>\text{rad}(a\cdot b\cdot c)^{1+\varepsilon}$ is finite.

It is known that $E_0$ is infinite. Now, computations suggests that also 
$\{(a,b,c)\in E_0|\exists p\in\mathbb P\exists k>1:b=p^{k}\}$ is infinite, where $\mathbb P$ is the set of primes, which would be a consequence of the conjecture, reformulated due to a suggestion of igael:  

Given a prime $p$, there exist a natural number $n$ and a natural number $a>0$, coprime to $p$ such that $a<p^n$ and $a+p^n\gt\operatorname{rad}(ap^n(a+p^n))=\operatorname{rad}(ap(a+p^n))$.  

Can this be proved? It holds for $k=3,\dots,100\,$ that it exists $a$ such that $a+p^4_k>\operatorname{rad}(ap^4_k(a+p^4_k))$. The conjecture also holds for $p=2,\,n=9$ and $p=3,\,n=5$.
If the conjecture is true, one can raise a reduced abc-conjecture for $a<b=p^n$.

There is a site with big files of unusual abc-tripples.
 A: Summary:  Tests made with the exact limitations above show no solution for $n=3$ and $p=41$ or $p=43$. Then the original conjecture is not valid "as is". 
However, if $n$ is taken as a free variable, perhaps we may always find solutions. For example, the $(a,p,n) \in \{ (49,41,5) , (49,43,5) , (1629,43,7) , (679,41,7) \}$ satisfy the inequality:
$$ a+p^n \gt \operatorname{rad}(a p^n (a+p^n)) $$ 

Discussion: The unsolved ABC conjecture states :

ABC Conjecture. For every $\varepsilon \gt 0$, there exist only finitely many triples $(A, B, C)$ of coprime positive integers with $A + B = C$, such
  that: 
$$ C \gt \operatorname{rad}(ABC)^{1+\varepsilon} $$

The original form of the Question conjectures that an infinite number of "near counter-examples" could be found: For any prime power $p^n$, $p$ an odd prime and odd $n \ge 3$, there exists $0 \lt a \lt p^n$ and coprime to $p$ such that:
$$ a+p^n \gt \operatorname{rad}(ap^n(a+p^n))$$
We recognize that these assignments:
$$ A=a , B= p^n, C = a+p^n $$
would have the current conjecture contradict the ABC-conjecture if $\varepsilon = 0$ (which of course is not allowed) because it asserts infinitely often (at least once for each odd prime $p$ and each odd $n\ge 3$):
$$ C \gt \operatorname{rad}(ABC) $$
Thus with $\varepsilon=0$, we allow ourselves to say "near counter-example".  However other constructions that achieve $C$ an arbitrarily large multiple of $\operatorname{rad}(ABC)$ are already known (see Section Examples of triples with small radical for details), though perhaps not of the exact form described in the Question.
Since $a$ can be any positive integer coprime to $p$ less than $p^n$, i.e. any of $p^{n-1}(p-1)$ choices, there are increasing opportunities for the proposed inequality to be satisfied as $p^n$ grows.  Let's give a heuristic argument for why it is not difficult to find such solutions.
If $A=a$ and $B=p^n$ are coprime, automatically $C=a+p^n$ is coprime to both $A,B$.  Thus $\operatorname{rad}(ABC) = \operatorname{rad}(A)\cdot \operatorname{rad}(B)\cdot \operatorname{rad}(C)$, and $\operatorname{rad}(B)=p$.  What we want to arrange is:
$$ \frac{a+p^n}{\operatorname{rad}(a+p^n)} \gt \operatorname{rad}(a)\cdot p $$
To make the left hand side "big", we want $a+p^n$ to contain either the square of a large factor or a high power of a small factor.  Again, we have $p^{n-1}(p-1)$ possible $a$ to choose from.
To make the right hand side "little", we can try choosing $a$ to be a very smooth number, one that has many small redundant prime factors.  For example, if $a=2^k$, then $\operatorname{rad}(a) = 2$.  The multiplier $\operatorname{rad}(p^n) = p$ is fixed for the purpose of our construction, but since $n \gt 1$, we know at least $p \lt p^n$.
Since the OP has added the restriction $a \lt p^n$, we restrict ourselves to choosing coprime residues $a\bmod p^n$ until we hit upon some $a+p^n$ containing a square factor $s^2$ with $s$ greater than $\operatorname{rad}(a)\cdot p$.  Then we would satisfy the current conjecture's inequality, since:
$$ s^2 \mid a+p^n \implies \frac{a+p^n}{\operatorname{rad}(a+p^n)} \ge s \gt \operatorname{rad}(a)\cdot p $$
Note on the algorithm used to check :
It is a brute force algorithm. For each $B$ and a maximal $n$, it computes $B^n$. Aware of overflow, it loops on A from 1 to B , computes the radical of abc and checks the inequality. ( Source available on request , a little big for the page format because it contains a very long list of primes )
Tests and output example :


*

*$5 + 28561 = 28566$ , $B = 13$ , $\text{rad}(4079367630) = 8970$ , $n = 4$ | $xto = 2,3,5,13,23$

*$B=17$ , $break=9321961974893840$ -> overflow, no solution found

*$1 + 6859 = 6860$ , $B = 19$ , $\text{rad} rad(47052740) = 1330$ , $n = 3$ | $xto = 2,5,7,19$
