How do I replace a D Flip Flop with a JK Flip Flop
Just as a hint, the general way to solve a problem like this is to write a truth table. The "inputs" to the truth table are the inputs to the machine plus the states of the flip-flops (one bit per FF), and the "outputs" are the outputs of the machine and the outputs of the flip-flops.

From that truth table, you can see if it's a Mealy machine. And given the desired "state on this time" and "state next time", and you can work out what the inputs to a J-K flip-flop should be:

 Q Q' | J K
------+-----
 0 0  | 0 X
 0 1  | 1 X
 1 0  | X 1
 1 1  | X 0


Where X is "don't care".

I only mentioned the truth table because you wanted to show that it was a Mealy machine, and finding the mapping of inputs and outputs is a definite way to get there.

As for the main question you're asking, you can't replace a D flip flop with a JK flip flop in the general case due to timing. D flip flops "read" their input on (say) the leading edge of the clock, and "output" their result effectively straight away, after a short propagation delay. JK flip flops, on the other hand, "read" on the leading clock edge and "output" on the trailing edge.

This property of JK flip flops can be useful; they are extremely resistant to combinational hazards, for example. But it does mean that they respond half a clock cycle later than a D flip flop.

Assuming that this isn't an issue (and I doubt it is in your circuit), you can just feed D into J and not-D into K.