Did you learn how to subtract?

Let the two randomly chosen irrational numbers be $\alpha = \overline{0.a_1a_2a_3 \cdots a_{2014}}$ and $\beta = \overline{0.b_1b_2b_3 \cdots b_{2014}}$ where $a_i,b_i \in \{0,1,2, \cdots ,9\}$ for $1 \leq i \leq 2014$ . Suppose $\alpha > \beta$ . For subtraction of $\beta$ from $\alpha$ without borrowing, $a_i \geq b_i$ .

Therefore, for $1 \leq i \leq 2014$ , $a_i \in \{b_i,b_i+1, \cdots ,9\}$ . Thus after choosing $b_i$ , $a_i$ can be chosen in $(10 - b_i)$ ways. So the number of favourable cases, given any $b_i$ , is $\sum_{j=0}^{9} (10 - j) = \sum_{j=1}^{10} j = 55$ . The total number of ways of selection of $a_i$ and $b_i$ is $10 \times 10 = 100$ .

Thus the required probability is (for $2014$ digits) $= (\frac{55}{100})^{2014}$ . Thus $100(p+k) = 100(0.55 + 2014) = \boxed{201455}$ . $\blacksquare$