When soap bubble is being inflated and its temperature remains constant, then it follows Boyle's law, so

pV = constant (k)

$$ \Rightarrow p = {k \over V}$$

Differentiating above equation with time, we get

$${{dp} \over {dt}} = k.\,{d \over {dt}}\left( {{1 \over V}} \right)$$

$$ \Rightarrow {{dp} \over {dt}} = k\left( {{{ - 1} \over {{V^2}}}} \right).\,{{dV} \over {dt}}$$

It is given that, $${{dV} \over {dt}} = c$$ (a constant)

So, $${{dp} \over {dt}} = {{ - kc} \over {{V^2}}}$$ ..... (i)

Now, from $${{dV} \over {dt}} = c$$, we get

$$dV = cdt$$

or, $$\int {dV = \int {cdt} } $$ or, $$V = ct$$ ..... (ii)

From Eqs. (i) and (ii), we get

$${{dp} \over {dt}} = {{ - kc} \over {{c^2}{t^2}}}$$

or, $${{dp} \over {dt}} = - {\left( {{k \over c}} \right)^{{t^{ - 2}}}}$$

$$ \Rightarrow dp = - {k \over c}.\,{t^{ - 2}}dt$$

Integrating both sides, we get

$$\int {dp = - {k \over c}\int {{t^{ - 2}}dt} } $$

$$p = - {k \over c}.\,\left( {{{{t^{ - 2 + 1}}} \over { - 2 + 1}}} \right)$$

$$ = - {k \over c}.\,{{ - 1} \over t} = {k \over {ct}}$$

or, $$p \propto {1 \over t}$$

Hence, p versus $${1 \over t}$$ graph is a straight line, which is correctly represented in option (b).