What is the effect of reducing NFFT less than the signal length
Reducing the NFFT samples in your DFT will result is less resolution in the spectrum you'll be observing. The frequency resolution is defined as 
$$
f_k = \frac{F_s}{N}k\quad \text{with}\quad k=0, 1, \ldots, N-1\tag{1}
$$
In your $N$-point Discrete Fourier Transform (DFT), decreasing $N$ increases the spacing between the $f_k$. Or put in another way, for a fixed sampling frequency $F_s$, 

\begin{align}
N_2 &amp;&lt; N_1 \implies \Delta f_{k_2} &gt; \Delta f_{k_1}\\
\text{Where}\quad f_{k_1} &amp;=\frac{F_s}{N_1}k_1\quad\text{and}\quad f_{k_2} =\frac{F_s}{N_2}k_2\\
\implies \Delta f_{k_1} &amp;=\frac{F_s}{N_1}\quad\text{and}\quad \Delta f_{k_2} =\frac{F_s}{N_2}
\end{align}

So say, you have some frequency components (i.e. peaks) you're observing in your spectrum with a high $N$, gradually reducing this value, you'll see that the peaks are getting less and less localized visually. Note that it is only visually, your DFT got your spectrum, you play with $N$ (i.e. NFFT) to visualise more out of your spectrum. Together with $F_s$, it determines the frequency resolution as defined in $(1)$.

EDIT:

To compute the energy, you sum the squared amplitudes spectrum values. Look for instance at the MATLAB code below for  the sine wave
$$
s(t) = 3\sin(2\pi 150t)
$$

Fs = 500;            % Sampling frequency
T = 1/Fs;             % Sampling period
L = 2000;             % Length of signal
t = (0:L-1)*T;        % Time vector
f = Fs*(0:(L/2))/L;

S = 3*sin(2*pi*150*t);

NFFT = 2^nextpow2(L);
NFFT1 = L;
NFFT2 = NFFT/2;
NFFT3 = NFFT/4;

Y = fft(S, NFFT1)/NFFT1;
Y2 = fft(S, NFFT2)/NFFT2;
Y3 = fft(S, NFFT3)/NFFT3;

EY = sum(Y.*conj(Y));
EY2 = sum(Y2.*conj(Y2));
EY3 = sum(Y3.*conj(Y3));


Your FFT have to be normalized and the energy computed as sum of the squared amplitude spectrum. For the sample code above you get:

EY = 4.5    
EY2 = 4.49644474953643  
EY3 =4.49832143842713


Which is from the amplitude of the sine $3$, giving ($3^2/2$). Also showing the effect of taking different $N$ values for the FFT on resulting energy.