The synthesis of 1-phenylprop-1-ene from propiophenone I would like to know how to make 1-phenylprop-1-ene ($\ce{C6H5-CH=CH-CH3}$) from propiophenone ($\ce{C6H5-CO-CH2-CH3}$)?

I think carbonyl group ($\ce{C=O}$) of propiophenone is electron-withdrawing group because of Oxygen of carbonyl group.
So the carbonyl group of propiophenone will be attacked by nucleophile. And this will be oxidized to make double bonds.

But I don’t know the detail reaction mechanism to make phenylprop-1-ene from propiophenone.
Can anyone help me understand this reaction mechanism?
This can be done it 2 simple steps. You are not adding to the carbon count so all you need to do is operate on propiophenone. 

Step 1 - reduce the carbonyl group to a secondary alcohol. Sodium borohydride in $\ce{MeOH}$ or $\ce{EtOH}$ will do this in high yield and the workup is straightforward.

Step 2 - dehydrate the secondary alcohol to give the double bond. As it is a benzylic alcohol, protonation of the $\ce{-OH}$ group and formation of the benzylic cation is strongly favoured and will give you the alkene you want. Catalytic para-toluenesulfonic acid in refluxing toluene (with water removal by Dean-Stark apparatus) will do this easily.