Answer:
\[KE\propto {{P}^{2}}\] \[\Rightarrow \frac{K{{E}_{2}}}{K{{E}_{1}}}=\frac{P_{2}^{2}}{P_{1}^{2}}\] \[\Rightarrow K{{E}_{2}}={{\left( \frac{2P}{P} \right)}^{2}}\times K{{E}_{1}}\] \[=4\,K{{E}_{1}}\] The kinetic energy becomes 4 times the initial.
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