Basil needs to solve an exercise on summing two fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$, where  $a$, $b$, $c$, $d$ are some non-zero real numbers. But instead of summing he performed multiplication  (correctly). It appears that Basil's answer coincides with the correct answer to given exercise. Find the value of  $\dfrac{b}{a} + \dfrac{d}{c}$.
1. **Given Problem**: Basil needs to sum two fractions $\frac{a}{b}$ and $\frac{c}{d}$, but he mistakenly multiplies them. The problem states that the result of his multiplication coincides with the correct sum of the fractions.

2. **Correct Sum of Fractions**: The correct sum of the fractions $\frac{a}{b}$ and $\frac{c}{d}$ is:
   \[
   \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}
   \]

3. **Incorrect Multiplication**: Basil's multiplication of the fractions is:
   \[
   \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}
   \]

4. **Equality Condition**: According to the problem, the result of the multiplication coincides with the correct sum:
   \[
   \frac{ad + bc}{bd} = \frac{ac}{bd}
   \]

5. **Simplifying the Equality**: Since the denominators are the same, we can equate the numerators:
   \[
   ad + bc = ac
   \]

6. **Rearranging the Equation**: To find the relationship between the variables, we rearrange the equation:
   \[
   ad + bc = ac \implies ad + bc - ac = 0 \implies ad - ac + bc = 0 \implies a(d - c) + bc = 0
   \]

7. **Solving for $\frac{b}{a} + \frac{d}{c}$**: We need to find the value of $\frac{b}{a} + \frac{d}{c}$. From the equation $a(d - c) + bc = 0$, we can isolate $bc$:
   \[
   a(d - c) = -bc \implies \frac{a(d - c)}{ac} = -\frac{bc}{ac} \implies \frac{d - c}{c} = -\frac{b}{a}
   \]

8. **Expressing $\frac{d}{c}$**: Adding $\frac{d}{c}$ to both sides:
   \[
   \frac{d}{c} - 1 = -\frac{b}{a} \implies \frac{d}{c} = 1 - \frac{b}{a}
   \]

9. **Summing the Fractions**: Now, we sum $\frac{b}{a}$ and $\frac{d}{c}$:
   \[
   \frac{b}{a} + \frac{d}{c} = \frac{b}{a} + \left(1 - \frac{b}{a}\right) = \frac{b}{a} + 1 - \frac{b}{a} = 1
   \]

Conclusion:
\[
\boxed{1}
\]