(A) In a series combination,the equivalent resistance $R_{s} = R_{1} + R_{2}$ is greater than the individual resistances. In a parallel combination,the equivalent resistance $R_{p} = (R_{1}R_{2}) / (R_{1} + R_{2})$ is smaller than the individual resistances.
$1$. For the $V-I$ graph (top): The slope of the graph represents resistance $(R = V/I)$. Since $R_{s} > R_{p}$,the line for 'series' must have a steeper slope than the line for 'parallel'. Thus,the top graph is correctly labelled.
$2$. For the $I-V$ graph (bottom): The slope of the graph represents conductance $(1/R = I/V)$. Since $R_{s} > R_{p}$,the conductance for parallel $(1/R_{p})$ is greater than for series $(1/R_{s})$. Therefore,the line for 'parallel' must have a steeper slope than the line for 'series'. Thus,the bottom graph is also correctly labelled.
Conclusion: Both graphs are correctly labelled.